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

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10
votes
1answer
3k views

Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
13
votes
1answer
1k views

Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?

Why isn't an infinite direct product of copies of $\Bbb Z$ a free module? Actually I was asked to show that it's not projective, but as $\Bbb{Z}$ is a PID, so it suffices to show it's not free. ...
27
votes
5answers
4k views

Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$ Can you tell ...
8
votes
1answer
856 views

When $\operatorname{Hom}_{R}(M,N)$ is finitely generated as $\mathbb Z$-module or $R$-module?

Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is ...
16
votes
1answer
718 views

Alternative construction of the tensor product (or: pass this secret)

The paper Tensor products and bimorphisms by B. Banachewski and E. Nelson studies tensor products (defined by classifying bimorphisms) in concrete categories. It is quite interesting that their main ...
4
votes
1answer
176 views

Are all simple left modules over a simple left artinian ring isomorphic?

I've a basic question. $R$ is a simple left artinian ring. I want to show that all simple left $R$-modules are isomorphic. A simple $R$-module $M$ is isomorphic to $R/J$ where $J$ is a maximal ...
2
votes
1answer
176 views

for any ring $A$ the matrix ring $M_n(A)$ is simple if and only if $A$ is simple

Let integer $n\geq 1$. I have obtained that for any field $k$, the matrix ring $M_n(k)$ is simple, i.e., $M_n(k)$ contains no nonzero proper two sided ideals. Now I want to prove that: for any ring ...
17
votes
2answers
2k views

Is $A \times B$ the same as $A \oplus B$?

When $A, B$ are $K$-modules, then $A \times B$ is the same as $A \oplus B$. Let $A, B$ be two $K$-algebras, where $K$ is a field. Is $A \times B$ the same as $A \oplus B$? Thank you very much. ...
12
votes
1answer
1k views

Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?

If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a commutative ring, does it follow that $m \geq n$? This is obviously true for vector spaces over a field, but how would one show ...
6
votes
3answers
2k views

Permutation module of $S_n$

Let $G=S_n$ and let $V$ be the permutation module of $G$ with basis $\{x_1,\cdots,x_n\}.$ Let $\lambda, \mu \in \mathbb{C}$ to allow one to define a $\mathbb{C}G$-homomorphism $\rho:V \to V$ by ...
8
votes
3answers
717 views

If $M$ is an artinian module and $f$ : $M$ $\mapsto$ $M$ is an injective homomorphism, then $f$ is surjective

If $M$ is an artinian module and $f$ : $M$ $\mapsto$ $M$ is an injective homomorphism, then $f$ is surjective. I somehow found out that if we consider the module $\mathbb Z_{p^{\infty}}$ denoting ...
20
votes
4answers
8k views

Example of modules that are projective but not free; torsion-free but not free

Free modules are projective, and projective modules are direct summands of free modules. Are there examples of projective modules that are not free? (I know this is not possible for modules of ...
2
votes
2answers
128 views

A ring without the Invariant Basis Number property

I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0 The ...
7
votes
2answers
126 views

Finitely generated free modules

Let $A$ be a ring and consider the free modules $A^{\oplus n}$, $A^{\oplus k}$, with $n,k\in \mathbb{N}$. Can $A^{\oplus n}$ be isomorphic to $A^{\oplus k}$ if $k\neq n$? Thanks in advance for the ...
6
votes
2answers
2k views

Why is the localization of a commutative Noetherian ring still Noetherian?

This is an unproven proposition I've come across in multiple places. Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian. Why is this? ...
19
votes
5answers
3k views

Every $R$-module is free $\implies$ $R$ is a division ring

From Grillet's Abstract Algebra, section VIII.5. Definitions. A division ring is a ring with identity in which every nonzero element is a unit. A vector space is a unital module over a division ...
15
votes
2answers
3k views

Example of a finitely generated module with submodules that are not finitely generated

I'm looking for an example of a finitely generated module with submodules that are not finitely generated. I've found a similar question dealing with group (i.e. an example of a finitely generated ...
12
votes
1answer
1k views

Hom of finitely generated modules over a noetherian ring

This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is ...
8
votes
2answers
238 views

Defining algebras over noncommutative rings

A definition of "algebra" (that is, an associative algebra, in the sense of ring theory) generally requires a commutative base ring. But there are cases where it's reasonable to consider algebras ...
8
votes
1answer
178 views

Is $R/N(R)$ a faithfully flat $R$-module?

I'm studying recently faithfully flat modules and I'd like to know the following: Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of ...
5
votes
2answers
817 views

Isomorphism between quotient modules

Is it true for a commutative ring $R$ and its ideals $I$ and $J$ that if the quotient $R$-modules $R/I$ and $R/J$ are isomorphic then $I=J$?
4
votes
2answers
139 views

Why an R-module is an R/I-module precisely when it is annihilated by I?

Let $I$ be an ideal of $R$ and let $M$ be an $R$-module. Prove that the formula $(r + I) \cdot m = r \cdot m$ for $r + I \in R/I$, and $m \in M$ makes $M$ into an $R/I$-module if and only if $i ...
2
votes
1answer
137 views

Lemma 1.3.4(b) in Bruns and Herzog

My question refers to the proof of the second of the following lemma given in Cohen-Macaulay rings by Bruns and Herzog. Lemma 1.3.4 (Bruns and Herzog): Let $(R,\mathfrak m,k)$ be a local ring, and ...
1
vote
1answer
152 views

A relation involving an endomorphism of a finitely generated module

Suppose $R$ is a commutative ring, $I$ is an ideal of $R$, and $M$ a finitely generated $R$-module. (Usually in this problem $R$ includes $1_R$.) Let $\phi : M \to M$ be an $R$-homomorphism, and ...
19
votes
2answers
1k views

Examples proving why the tensor product does not distribute over direct products.

I recently read about the result that the tensor product distributes over direct sums. I was curious if it also distributes over direct products, but google tells me it doesn't. What are some simple ...
9
votes
1answer
162 views

Infinite linear independent family in a finitely generated $A$-module

So I'm stuck with this problem. Let $A$ be a nonzero commutative ring (with unit). I have several questions that are really close to each other. 1) Let $M$ be a finitely generated module over $A$. ...
6
votes
3answers
670 views

$R^n \cong R^m$ iff $n=m$

How can i show that two $R$-modules of finite rank are isomorphic if and only if they have the same rank, i.e., $R^n \cong R^m$ iff $n=m$.
9
votes
3answers
2k views

If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. [duplicate]

Possible Duplicate: Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field I need to prove this result, but the only starting point I think of is to ...
3
votes
2answers
1k views

Finitely generated modules in exact sequence

For $A$-modules and homomorphisms $0\to M'\stackrel{u}{\to}M\stackrel{v}{\to}M''\to 0$ is exact. Prove if $M'$ and $M''$ are fintely generated then $M$ is finitely generated.
6
votes
2answers
352 views

Isomorphism of an endomorphism ring, how can $R\cong R^2$?

I'm aware that over a commutative ring, two free finitely generated modules of finite rank are isomorphic if and only if they have the same rank. I'm trying to understand a curious example of why ...
4
votes
2answers
321 views

Atiyah and Macdonald, Proposition 2.9

The following simple claim is used without proof in Proposition 2.9 of Atiyah and MacDonald (p.23). Although I believe I can prove it with a fairly involved argument, the claim is treated by the ...
2
votes
1answer
124 views

$R/Ra$ is an injective module over itself

Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself. If $R$ is a PID, every $R$- divisible module is injective, but the question concerns ...
5
votes
1answer
338 views

Injective Cogenerators in the Category of Modules over a Noetherian Ring

Let $R$ be a Noetherian ring and let $\mathcal{A}$ be an injective $R$-module. The injectivity of $\mathcal{A}$ is equivalent to the exactness of the functor $Hom_R(-,\mathcal{A})$, i.e. whenever we ...
3
votes
1answer
80 views

Coker of powers of an endomorphism

Let $F\in\operatorname{End}_R(M)$, where $M$ is a Noetherian $R$-module. If $\operatorname{Coker}F$ is of finite length, is Coker and Ker of all powers of $F$ of finite length? Is the condition of ...
2
votes
1answer
210 views

Artinian ring and faithful module of finite length

Let $A$ be a ring. How can I prove that: $A$ is an Artinian ring $\Leftrightarrow \exists$ a faithful $A$-module which is of finite length. I know that if a ring has a faithful $A$-module which ...
1
vote
2answers
130 views

Endomorphisms of modules satisfying chain conditions and counterexamples.

1) How can I show that any one to one endomorphism of an Artinian module is an automorphism. 2) I also want to show that any onto endomorphism of a Noetherian module is an automorphism. I need ...
25
votes
4answers
1k views

Why isn't $\mathbb{C}[x,y,z]/(xz-y)$ a flat $\mathbb{C}[x,y]$-module

Why isn't $M = \mathbb{C}[x,y,z]/(xz-y)$ a flat $R = \mathbb{C}[x,y]$-module? The reason given on the book is "the surface defined by $y-xz$ doesn't lie flat on the $(x,y)$-plane". But I don't ...
25
votes
4answers
754 views

Pathologies in module theory

Linear algebra is a very well-behaved part of mathematics. Soon after you have mastered the basics you got a good feeling for what kind of statements should be true -- even if you are not familiar ...
23
votes
2answers
5k views

Proving that the tensor product is right exact

Let $A\stackrel{\alpha}{\rightarrow}B\stackrel{\beta}{\rightarrow}C\rightarrow 0$ a exact sequence of left $R$-modules and $M$ a left $R$-module ($R$ any ring). I am trying to prove that ...
20
votes
4answers
2k views

Reference request: compact objects in R-Mod are precisely the finitely-presented modules?

Let $R$ be a ring. According to this MO question, the modules $M \in R\text{-Mod}$ such that $\text{Hom}(M, -)$ preserves all filtered colimits (the compact objects) are precisely the ...
10
votes
2answers
875 views

Elementary proof that if $A$ is a matrix map from $\mathbb{Z}^m$ to $\mathbb Z^n$, then the map is surjective iff the gcd of maximal minors is $1$

I am trying to find an elementary proof that if $\phi$ is a linear map from $\mathbb{Z}^n\rightarrow \mathbb{Z}^m$ represented by an $m \times n$ matrix $A$, then the map is surjective iff the gcd ...
7
votes
4answers
768 views

Ring of polynomials as a module over symmetric polynomials

Consider the ring of polynomials $\mathbb{k} [x_1, x_2, \ldots , x_n]$ as a module over the ring of symmetric polynomials $\Lambda_{\mathbb{k}}$. Is $\mathbb{k} [x_1, x_2, \ldots , x_n]$ a free ...
7
votes
3answers
1k views

Learning to think categorically (localization of rings and modules)

I've been reading Ravi Vakil's notes on algebraic geometry notes and am having a hard time connecting the standard definition of the localization of a module to the definition in terms of universal ...
10
votes
2answers
2k views

Finitely generated projective module

Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free? What I know about a finitely generated projective module $M$ over a PID $R$ ...
9
votes
4answers
310 views

The connection between decomposition of $\ker(f_\phi g_\phi)$ and the Chinese remainder theorem

Original problem Suppose $K$ is a field, $f,g\in K[x]$ and $\gcd(f,g)=1$. $V$ is a linear space based on $K$ and $\phi\in\operatorname{End}(V)$. Notation $f_\phi$ and $g_\phi$ denote linear ...
8
votes
1answer
910 views

Projective and injective modules; direct sums and products

I need two counterexamples. First, a direct sum of $R$-modules is projective iff each one is projective. But I need an example to show that, “an arbitrary direct product of projective modules need ...
10
votes
1answer
905 views

Countable infinite direct product of $\mathbb{Z}$ modulo countable direct sum

Let $M=\mathbb Z^{\mathbb N}$ be the product of a countable number of copies of the group $\mathbb{Z}$ and let $N=\mathbb Z^{(\mathbb N)}$ be the direct sum of a countable number of copies of ...
8
votes
3answers
291 views

Is Orzech's generalization of the surjective-endomorphism-is-injective theorem correct?

In math.stackexchange answer #239445, Makoto Kato quoted a statement from the paper Morris Orzech, Onto Endomorphisms are Isomorphisms, Amer. Math. Monthly 78 (1971), 357--362. The statement ...
4
votes
1answer
239 views

A counterexample to the isomorphism $M^{*} \otimes M \rightarrow Hom_R( M,M)$

I am going over some counterexamples for the the isomorphism $M^{*} \otimes M \rightarrow Hom_R( M,M)$. In particular I have been trying to understand what happens if you remove the various ...
7
votes
6answers
2k views

Some basic book to start with modules?

I am very interested in studying modules, which I have studied algebra, is basic theory of groups and rings, of Hungerford and Dummit. I was reading the Dummit the module, but I still struggled a bit ...