For questions about modules over rings, concerning either their properties in general or regarding specific cases.

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The ring $R$ is a field iff $R$ has a finitely generated divisible module

I have the following problem: Prove that a integral domain $R$ is a field iff there exist a finitely generated divisible $R$-module. One side is easy, because if $R$ is a field then $R$ itself ...
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2answers
36 views

An integral domain and its field of fractions.

I'm trying to solve the following problem: Let $R$ be a integral domain which is not a field and $K$ its fractions field. Show that a non-zero module $R$-homomorphism from $K$ to $R$ does not ...
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0answers
61 views

$\operatorname{supp}(M) \subseteq \operatorname{supp}(N) \iff f_I(M)\subseteq f_I(N) $?

Let $ R $ be a commutative unital ring, $ I $ an ideal of $ R $, and $ M $ an $ R $-module. It has proven (here) that if $\operatorname{supp}(M) \subseteq \operatorname{supp}(N)$ then ...
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1answer
13 views

Torsion and module over a ring.

I need to solve the following problem: Show that a ring $R$ is a field iff every $R$-module is torsion free module. The "only if" part is quite easy because if $R$ is a field then every ...
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1answer
33 views

Characterization of right noetherian rings

Here's a quick question on noetherian rings. I know that for a ring $R$, the following are equivalent. $R$ is left noetherian Every finitely generated left $R$-module is noetherian Every submodule ...
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1answer
37 views

A two sided ideal of a Noetherian ring

$R$ is a left Noetherian ring with a minimal left ideal. Consider the set of minimal left ideals of $R$ ordered by inclusion. Then there is a maximal element $\mathfrak b= \bigoplus_{i\in I} \mathfrak ...
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3answers
46 views

Example of ring $R$ with ideals $I\neq J$ such that $R/I \cong R/J$ as modules

It's easy to prove that if $I$, $J$ are two-sided ideals and $R/I\cong R/J$ as modules over $R$, then $I=J$. What about left ideals? Is there a simple counterexample? I believe I've found an answer, ...
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1answer
42 views

Show that $D \cong {\rm End}_A(D^n)$ where $D$ is a division algebra and $A\cong M_n(D)$

Define $A$ to be a finite dimensional simple algebra over a field $k$. $D$ is a $k$-division algebra (not necessarily commutative) such that $A\cong M_n(D)$ for some integer $n$. Let $L$ be a minimal ...
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1answer
76 views

A not free $\mathbb{Z}$-module

From this post I see that $\mathbb{Z}^{\mathbb{N}}$ is not free. Someone can explain me why? Thanks a lot!
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1answer
21 views

ideal,ring,flat module,modules over R

Is there a characterization of modules (AND equivalent characterizations of rings R) over integral domains R with the property that each left ideal in R is flat?When all left ideals are ...
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1answer
52 views

A Direct Sum of Members of a Certain Class of Modules

Let $S$ be a class of $R$-modules and let an $R$-module $M$ be countably generated. Suppose that, for every direct summand $K$ of $M$, each element of $K$ belongs to a direct summand of $K$ that is ...
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1answer
49 views

A condition of equivalence of flatness and projectiveness

This is a problem in "Foundations of Module and Ring Theory" of Wisbauer: " Let $R$ be a subring of the ring $S$ containing the unit of $S$. Show that a flat $R$-module $N$ is projective if and only ...
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0answers
59 views

Surjectivity implies injectivity of finitely generated modules, localization?

The following problem is canonical: Suppose $A$ is a commutative unitary ring, and $M$ is a finitely generated module over $A$. If an endomorphism $f\colon M\to M$ is surjective, then it's also ...
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1answer
26 views

Simple $M_n(D)$-module with $D$ a division ring

Define $D$ to be a division algebra over a field $k$ and $R=M_n(D)$ the $n\times n$ matrix ring over $D$. A simple $R$-module $M$ is the quotient of $R$. I can write $R=\bigoplus_j I_j$ where $I_j$ is ...
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1answer
53 views

Confused about proof in Basic abstract algebra by Bhattacharya

On page 264 , 2nd edition. Theorem 5.1 It says Let M be a free $R$-module with "a basis" $\{e_1,\dots,e_n\}$ Then $M$ is $R$-isomorphic to $R^n$. Above he is defining the standard basis as the ...
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0answers
41 views

Simple Cogenerator for the category of left $R$-modules

I want to prove that if the category of left $R$-modules has a simple cogenerator $M$, then $R$ is a simple ring. (Here, $R$ is an arbitrary ring with $1$.) My try is to take an ideal $I$ of $R$ ...
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1answer
35 views

Submodules and quotients of free modules over Noetherian local rings

Let $R$ be a Noetherian local commutative ring, $F$ a finitely generated free $R$-module and $A,B$ some arbitrary $R$-modules. Consider a short exact sequence $0 \to A \to F \to B \to 0$. In [Bruns, ...
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1answer
91 views

A question about modules. [closed]

My Algebra book says that a module is the following map $$R\to End(M)$$ This implies that if $a,b\in M, a\neq b$, then $ra\neq rb$ for some $r\in R$. I dont see why that is.
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1answer
83 views

Example of strict inclusion for the localization of associated primes

Let $A$ be a commutative ring and $M$ an $A$-module. It is well known that $$\operatorname{Ass} M\cap\operatorname{Spec}S^{-1}A\subset\operatorname{Ass}S^{-1}M,$$ and that equality holds if $A$ (or ...
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Exercise 7.10 Atiyah, $M[x] $ is a noetherian $A[x] $-module [duplicate]

The exercise is: Let $M$ be a noetherian $A$-module. Then $M[x] $ is a noetherian $A[x] $ module. The action of $A[x] $ on $M[x] $ is the obvious one. In a previous exercise it was shown that ...
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0answers
29 views

Tensor product of $A_n$ modules/ localisation at ring of differentials

I'm working through Coutinho's "A Primer of Algebraic D-Modules" and I've gotten stuck on the following question: Let $p \in K[x_1, \ldots ,x_n]$ be non-zero, and let $A_n$ be the Weyl Algebra. Show ...
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2answers
42 views

$R^{(I)} \cong K \oplus H$ where $R^{(I)}$ is free but $K$ is not free

Let $R$ be a commutative ring with unit. Is there an example of a direct sum of $R$-modules $$R^{(I)} \cong K \oplus H$$ where $R^{(I)}$ is free but $K$ is not free ? Clearly $R$ can't be a PID.
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21 views

Classification of separable algebras over a commutative ring

A separable algebra over a field can be classified as a finite product of matrix algebras over division algebras whose centers are finite dimensional separable field extensions of the field. (See ...
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1answer
42 views

Kahler forms of a smooth affine algebra vanish eventually?

If $k$ is a Noetherian ring, then do the Kahler forms of a smooth affine $k$-algebra of dimension $d$ vanish above $d$? I mean is: $\Omega^{d+1}_{A|k}\cong 0$?
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26 views

How Can I prove that this module is isomorphic to the generators with relations…

Let $G$ be a finite group of order $n$. Let $R=\mathbb{Z}G$ and $N=\sum_{g\in G} g$, observe that $gN=Ng=N$, $N^2=nN$. Let $r\in \mathbb{Z}$ be prime to $n$ and let $P_r$ be the ideal of $R$ generated ...
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0answers
82 views

What is $\operatorname{Ass}\operatorname{Ext}^i(M,N)$?

This is exercise 1.2.27 of Bruns-Herzog: Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $N$ an arbitrary $R$-module. Deduce that $\operatorname{Ass}(\operatorname{Hom}_R(M,N)) = ...
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2answers
63 views

Decomposition of finitely generated graded modules over PID

I found this decomposition theorem used in a paper I'm reading, but it isn't referenced and I can't seem to find it in any of the books I have: Every graded module $M$ over a graded PID decomposes ...
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1answer
35 views

Submodules of semi-simple modules

Let $R$ be a ring (with unity, not necessarily commutative) and let $P$ be an irreducible $R$-module. Let $$M=\bigoplus_{i=1}^r P$$ be a direct sum of $r$ copies of $P$, for some $r\geq 1$. Then, $M$ ...
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0answers
71 views

Automorphism of certain f.g. free modules

This is a quick question from Frohlich and Taylor's Algebraic Number Theory, II.4, p 94. Let $R$ be a Dedekind domain with quotient field $K$, $\mathfrak p$ is a non-zero prime ideal of $R$ and ...
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1answer
32 views

Basis for the completion of a free module

This (or similar) question might have been asked before- apologies for any duplication. I've got a Dedekind domain $R$, a non-zero prime ideal $P$ of $R$ and the completion $\widehat{R}$ of $R$ wrt ...
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1answer
74 views

Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?

The question is contained in title, I'm working with $A$-modules $M$ and $N$. I feel like Yoneda's lemma is what I'm looking for but it applies to functors into the category of sets, whereas ...
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1answer
75 views

How to prove this innocent looking isomorphism

I've got a Dedekind domain $R$ with quotient field $K$, a non-zero prime ideal $P$ of $R$. I form the completion $\widehat{K}$ of $K$ wrt the valuation $v_P$ associated to $P$. Let $\widehat{R}$ be ...
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2answers
30 views

Why does tensoring a projective resolution with a flat module give another projective resolution?

This question came up in this thread: Proving that tensoring a projective module with a flat module gives a projective module? Assume $\left\{P_i\right\}$ is a projective resolution of an $R$-module ...
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68 views

Direct product of direct sum of a flat module

I have a problem concerning flat modules: Let $M$ be an $R$-module such that the direct product $M^A$ is flat for all sets $A$. I want to prove that $(M^{(B)})^A$ is also flat for any sets $A$ ...
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1answer
39 views

If $0$ is the zero-object $ \Longrightarrow F(0) $ is the zero object when $F$ additive

Let $$ F : \text{A-Mod} \to \text{A-mod} $$ be an additive functor. Then if $0$ is the zero-object $F(0) $ is the zero object. Why this is true ? The definition of additive functor that I know is ...
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1answer
45 views

Showing that $\bigoplus_{i\in\mathbb N}\mathbb Z/2\mathbb Z$ is not a direct summand of $\prod_{i\in\mathbb N}\mathbb Z/2\mathbb Z$

I know that if the direct sum of countably many copies of $\mathbb Z/2\mathbb Z$, $I$ were a direct summand of the direct product of countably many copies, $R$ (direct summand as $R$-modules), then ...
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2answers
125 views

Proving that tensoring a projective module with a flat module gives a projective module?

If $P$ is a projective module and $M$ is a flat module, both over some commutative ring $R$, then how do you prove that $M\otimes_R P$ is flat? I tried using the fact that $P$ is the direct ...
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1answer
26 views

Tensor product and free $R$-module construction

I am trying to understand an exposition on Tensor products by Keith Conrad. In the proof of Theorem 3.2 on page 7 it considers the free $R$-module on the set $M \times N$: $$F_R(M \times N) = ...
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1answer
56 views

Linear dual of vector fields

Suppose that $M$ is a smooth manifold and $\mathfrak{X}(M)$ is the set of smooth vector fields on $M$. There are basically two different linear structures on $\mathfrak{X}(M)$: 1.) $\mathfrak{X}(M)$ ...
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When is an $R$-projective module a projective module?

Let $R$ be a semiperfect ring. Is it a true fact that every $R$-projective module $M$ with $Rad(M)$ superfluous in $M$ is projective? I could not reach a good result using just the fact that ...
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2answers
47 views

Relation between $\operatorname{Coker}(f)$ and $\operatorname{Coker}(f \otimes 1_P)$

Let $M,N,P$ be $R$-modules ($R$ commutative ring with $1$) and let $f:M\to N$ be a $R$-module homormorphism. Let tensor the homomorphism to get $ f \otimes 1_P : M \otimes P \to N \otimes P $. I ...
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1answer
66 views

Completion of a torsion-free module

Let $R$ be a Dedekind domain, $K$ its field of fractions, $P$ a non-zero prime ideal of $R$. Let $\hat R_P$ be the completion of $R$ wrt to the valuation $v_P$ induced by $P$ and let $L$ be a ...
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1answer
62 views

Prove that if the induced homomorphism $M/\mathfrak aM \to N/\mathfrak aN$ is surjective, then $f$ it's surjective.

This problem it's from Atiyah and Macdonald, Chapter 2. Let $A$ be a commutative ring with $1 \ne 0$ and let $\mathfrak a$ be an ideal of $A$ contained in the Jacobson radical. Let $M$ be an ...
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1answer
24 views

Topological Tensor Product is a Topological Ring Independent of the Choice of Basis

Let $A, B$ be commutative rings containing a field $k$, with $B$ a finite dimensional $k$-module, $w_1, ... , w_N$ a basis. If $w_iw_j = \sum\limits_{n=1}^N c_{ijn}w_n$, then we can define $C$ to be ...
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49 views

Soft question (Etymology - Flatness)

Why where flat modules named "flat"? Is it because they are necessarily torsion free so in a "not convoluted" or circular like $\mathbb{Z}/n\mathbb{Z}$ is as a $\mathbb{Z}$-module?
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1answer
71 views

Why is the Ideal Norm Multiplicative?

I had asked this question before and got a partial answer. Let $A$ be a Dedekind domain with quotient field $K$, $L$ a finite separable extension of $K$ of degree $n$, and $B$ the integral closure of ...
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1answer
45 views

How can I prove commutative rings are stably finite?

I have a proof that proves that a stably finite ring (i.e. one where $AB=I\iff BA=I$ for any two square matrices $A,B$ with entries in the ring) has the Invariant Basis Number (IBN). Trouble is, it is ...
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1answer
44 views

Localization of a torsion-free module

I have a Dedekind domain $R$ with field of fractions $K$ and a non-zero prime ideal of $P$ of $R$. Let $L$ be a torsion-free $R$-module. How can I show that $R_P\otimes_R L$ is torsion-free as an ...
2
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1answer
39 views

When is $M \times N$ contained in$ M \otimes_{R} N$.

Let $R$ be a ring with 1. $M$ is a right R-module and $N$ is a left R-module. A tensor product of M and N comes with a map from $M \times N \to M\otimes N$ which is actually a composition of maps ...
2
votes
1answer
33 views

Two actions of $U(\mathfrak{h})$ on $U(\mathfrak{g})$ where $\mathfrak{h}\hookrightarrow\mathfrak{g}$

Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$, then by PBW theorem we know $U(\mathfrak{h})\hookrightarrow U(\mathfrak{g})$. Let $\{x_i, y_i\}$ be an ordered basis of $\mathfrak{g}$ where ...