The tag has no wiki summary.

learn more… | top users | synonyms

-1
votes
0answers
23 views

Property of free submodules for a module over a PID [migrated]

This question was asked here and remains without solution. It's possible to produce an example of an integral domain $R$ and a free $R$-module $M$ with free submodules $L, L'$ such that $L+L'$ is ...
-1
votes
1answer
90 views

behavior of multiplicity in exact sequences

Bruns-Herzog define multiplicity When the ring and module are not necessarily graded as $e(M)=e(gr_m(M))$, see B-H 4.6. I have two questions: Question1. Many concepts in commutative algebra have ...
0
votes
0answers
90 views

Derived categories of modules categories

Does anyone know if there is a note or a paper about the derived category of the category $\sigma[M]$ where $M$ is a left module over a ring?, and some uses of this.
3
votes
1answer
82 views

When are countably generated Hilbert modules generated by c.p.c. order zero maps?

Throughout let $B$ be a stable C*-algebra, i.e. $B\cong B\otimes K$, where $K$ is the C*-algebra of compact operators on an infinite dimensional separable Hilbert space. It is well-known that any ...
0
votes
1answer
122 views

Commutation of tensor products with inverse limits in a specific case

For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative ...
3
votes
1answer
84 views

When does a faithful module have an element with zero annihilator?

This is a follow up of Example of a finitely generated faithful torsion module over a commutative ring. Let $M$ be a finitely generated module over a commutative ring $R$ with the property that ...
-1
votes
1answer
79 views

Is this apushout diagram [closed]

Let $A, B, C, E$ and $F$ be some objects in an abeleian category $\mathcal{C}$. Let we have a commutative diagram \begin{array}{ccccccccc} 0 & \xrightarrow{} & A & \xrightarrow{f} & ...
1
vote
0answers
74 views

A noncommutative vector bundle associated with a codimension one foliation

Assume that we have a codimension one foliation of a manifold $M$ which is generated by a one form $\alpha$. So the following $\phi$ satisfies $\phi \circ \phi =0$:$$\phi:\Omega^{i}(M)\to ...
1
vote
1answer
109 views

Annihilator of tensor product when $R$ is domain

Let $R$ be a Noetherian domain and $M$ and $N$ be two faithful $R$-modules. Is it true that $\operatorname{Ann}_R(M\otimes_R N)=0$?
1
vote
1answer
68 views

$IM=mM$. can we say that $I$ is a reduction ideal of $m$?

Question. Let $(R,m)$ be a Noetherian local ring and $M$ be a finite faithful $R$-module. Let $I$ be an ideal of $R$ such that $IM=mM$. Can we say that $I$ is a reduction ideal of $m$? Recall that $I$ ...
3
votes
3answers
162 views

canonical module can be identified with an ideal. how can one reach that ideal?

Let $R[[X,Y,Z]]/(X,Y)\cap (Y,Z)\cap(X,Z)$. then $R$ is Cohen-Macaulay ring and has a canonical module, $K$. By Proposition 3.3.18 of Bruns_Herzog, $K$ can be identified with an ideal in $R$. So we ...
1
vote
1answer
110 views

Graded version of Baer's Criterion

Baer's Criterion for injectiveness of modules says: "An $R$-module $E$ is injective iff for all ideals $I$ of $R$, every homomorphism $f\colon I \to E$ can be extended to $R$." I wonder if there is a ...
2
votes
1answer
198 views

When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?

Let $\mathrm{Quillen}$ be the model category of simplicial sets with the Quillen model structure, and $\mathrm{Joyal}$ the model category of simplicial sets with the Joyal model structure. As is ...
4
votes
1answer
252 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 ...
0
votes
0answers
64 views

An embedding of modules by tensor product over a Noetherian domain

I have a problem on Ring theory. I would like to prove or disprove the following statement: Let $R$ be a Noetherian domain. Then by the Goldie theorem $R$ have $Q$ as a full ring of quotients and $Q$ ...
4
votes
0answers
82 views

Multiplicity of $Ext^{d-t}(M,\omega_R)$, ($d=\dim R, t=\dim M$)

Let $R=\bigoplus_{i \geq 0} R_i$ be a Cohen-Macaulay graded ring ($R_0$ is a field and $R$ is generated by $R_1$) of dimension $d$ with canonical module $\omega_R$, and $M$ a graded Cohen-Macaulay ...
3
votes
1answer
89 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 ...
6
votes
0answers
147 views

Correspondences as generalized morphism between $C^*$-algebras

While reading about Morita equivalence in the category of $C^*$-algebras I met also the following notion: a correspondence between two $C^*$-algebras $A,B$ is a pair $(X,\varphi)$ where $X$ is a ...
5
votes
1answer
113 views

Isomorphism of Hilbert modules

In Hilbert space theory if we have a linear bijection $U:H \to K$ which is an isometry, then it preserves inner products. Suppose now that you have two (right) Hilbert modules $X,Y$ (say, over the ...
-1
votes
1answer
141 views

Ext functor for more than two modules? [closed]

The question is natural. Let's just work in the category of modules over a ring. Pick three modules $M_1, M_2, M_3$. Consider consecutive extensions of these modules, i.e., consider M, such that we ...
2
votes
1answer
237 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 ...
2
votes
1answer
145 views

what are the possible approximations for ideals

(Fix some local ring $(R,\mathfrak{m})$ over a field of zero characteristic.) Suppose an ideal $J$ is defined by some complicated formula/procedure. And there is no hope of computing it/or writing ...
0
votes
1answer
60 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 ...
0
votes
1answer
217 views

Uniform Artin-Rees

The Artin-Rees lemma states that if $R$ is a Noetherian ring, $I \subseteq R$ is an ideal and $N \subseteq M$ are finitely generated $R$-modules, then there exists an integer $k$ such that for every ...
1
vote
1answer
137 views

Lifting a direct summand of a free module

[EDIT]: After getting a nice counter example provided by Steven Landsburg I realize that I forgot to impose an important condition...namely $R$ is supposed to be complete w.r.t. the $I$-adic topology. ...
0
votes
0answers
42 views

How to generalize balanced and absorbing sets to R-modules?

I'm looking for generalizing the notions of balanced set and absorbing set. The goal is using them for analyzing topological R-modules with R being a unit ring. It's easy to generalize balanced and ...
3
votes
1answer
99 views

Let $A\in\operatorname{M}_n(F)$ be a matrix, how to prove $\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}$

I asked the following question on Math Stack Exchange, but no people reply. I know MO is more professional and it is for mathematicians to discuss research problems. Maybe this question is unsuitable ...
1
vote
0answers
49 views

a generalization of the annihilator of cokernel ideal

Let $R$ be a (commutative, associative, unital) ring, consider a homomorphism of some (finitely generated) free $R$-modules $F\stackrel{A}{\rightarrow}G$. Its basic invariants are the Fitting ideals, ...
9
votes
1answer
529 views

Is $\mathbb{R}$ a $\mathbb{C}$-module without AC?

Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial(scaler multiplication is not identicaly zero) $\mathbb{C}$-module. Now my questions are? 0.Is it consistent with $ZF$ that $\mathbb{R}$ is ...
4
votes
2answers
231 views

Canonical presentation of pro-modules over pro-rings

Let $A = (\dotsc \twoheadrightarrow A_2 \twoheadrightarrow A_1 \twoheadrightarrow A_0)$ be a (commutative) pro-ring with surjective transition maps. Consider the category $\mathcal{M} := \varprojlim_i ...
1
vote
0answers
106 views

Ring of endomorphisms as a criterion of a dimension 1 module

Let $R$ be a unital ring and $M$ be an $R$-module. I have some questions about relation between the ring $\operatorname{End}_R M$ of endomorphisms and the notion of “dimension” of a module. I’m not ...
1
vote
0answers
440 views

Submodule embeddable in a finitely generated module

I have a terminology question. For a commutative ring $A$ (not necessarily Noetherian), $A$-modules that are isomorphic to an $A$-submodule of a finitely generated $A$-module form a fairly good class ...
5
votes
1answer
234 views

Does $ \text{mult}(R / I) = d_{1} \cdots d_{r} $ imply that $ (f_{1},\ldots,f_{r}) $ is an $ R $-regular sequence?

We define the multiplicity of an $ R $-module $ M $ of dimension $ d > 0 $ to be $$ \text{mult}(M) \stackrel{\text{df}}{=} \text{LC}(P_{M}) \cdot (d - 1)!, $$ where $ P_{M} $ denotes the Hilbert ...
5
votes
1answer
284 views

Is a projective module of constant finite rank finitely generated?

If $R$ is a (commutative) ring and $P$ is a projective $R$-module, then every localization of $P$ at a prime of $R$ is free by Kaplansky's theorem, and has a well-defined rank. If these ranks are all ...
4
votes
0answers
55 views

Is there a positive integer k such that any endomorphism of any free module over any commutative ring is a linear combination of k idempotents?

Consider the following Condition (C) on a positive integer $k$: (C) If $R$ is a commutative ring, if $F$ is a free $R$-module, and if $f$ is an endomorphism of $F$, then $f$ is an $R$-linear ...
-3
votes
1answer
113 views

Isomorphic quotient of a Module over Noetherian commutative algebra [closed]

I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve ...
2
votes
1answer
124 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 ...
2
votes
0answers
54 views

an equality in module theory

Let $R$ be a ring and $M$ be an $R$-module. Let $N$ and $L$ be two submodules of $M$ such that $(rad(N+L):M)=\sqrt{(N:M)+(L:M)}$. Then can we conclude that $(N+L:M)=(N:M)+(L:M)$? Note that ...
1
vote
0answers
210 views

R[[X]] flat as a R[X]-module?

I assume $R[X]\rightarrow R[[X]]$ is not flat in general, but I was wondering if any conditions on a commutative ring $R$ are known such that $R[[X]]$ is flat as a $R[X]$-module. Would $R$ noetherian ...
6
votes
0answers
95 views

Finiteness for separated residually finite modules

Suppose that $A$ is a commutative noetherian Jacobson ring and $M$ is an $A$-module. Suppose in addition that $M$ is $\mathfrak{m}$-adically separated for every maximal ideal $\mathfrak{m}$, and that ...
5
votes
1answer
192 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. ...
1
vote
2answers
281 views

A semisimple group ring

Let $n \in \mathbb{N}$, $p$ a prime number, and $G$ a finite group of order coprime to $p$. Let $R = \mathbb{Z} /p^n \mathbb{Z}$ be the ring of integers mod $p^n$. Must $R[G]$ be semisimple? As noted ...
3
votes
0answers
375 views

An exact sequence which does not split

Let $X$ and $Y$ be indecomposable modules over a finite dimensional algebra and let $f \colon X \to Y$ be a non-zero morphism which is neither a monomorphism nor an epimorphism. Suppose that it is ...
2
votes
1answer
62 views

dg-flat complexes and their characters

Let $\otimes$ denotes the usual tensor products of complexes and symbols live in the category of chain complexes of $R$-modules. Let $X$ be a dg-flat complex (i.e. $X_n$ is flat for each n and ...
-2
votes
1answer
260 views

Doubt in this proof of Horrocks theorem

I'm beginning to study some research papers and I need right now to understand the solution of Vaseršteĭn of Serre's theorem (simplest proof of this theorem), to do so, I'm beginning to understand ...
11
votes
7answers
729 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 ...
7
votes
1answer
154 views

Complexity of rational $\mathrm{GL}_{n(r)}$-modules

Let $k$ be an algebraically closed field of characteristic $p>0$, and let $G=\mathrm{GL}_n(k)$ for some natural number $n$. For any integer $r\ge 1$, let $G_{(r)}$ denote the $r$th Frobenius ...
1
vote
0answers
105 views

Intuitive idea: What is the motivation for relative homological algebra

Basically I was wondering what is the purpous of choosing a smaller class of epis in a category? Does it allow for the existence of "more" "projective-like" objects? What are some applications? For ...
4
votes
1answer
366 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 ...
10
votes
1answer
676 views

Why are they called Specht Modules?

I know that the simple modules of $\mathbb{C}S_n$ are called Specht Modules, and they are named after the German Mathematician Wilhelm Specht because he studied them, but I think these modules were ...