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

learn more… | top users | synonyms

22
votes
4answers
2k 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
684 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
624 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
2k 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 ...
2
votes
1answer
147 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 ...
14
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. ...
9
votes
1answer
876 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. ...
10
votes
1answer
965 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 ...
16
votes
4answers
6k 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 ...
4
votes
2answers
1k 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? ...
18
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 ...
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 ...
7
votes
2answers
172 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 ...
6
votes
3answers
1k 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 ...
4
votes
1answer
107 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 ...
8
votes
1answer
144 views

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

So I'm stuck with this problem. Let $A$ be a commutative ring (with unit). I have several questions that are really close to each other. 1) Let $M$ be a finitely generated module over $A$. Can we ...
8
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 ...
2
votes
1answer
83 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 ...
7
votes
3answers
497 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 ...
6
votes
2answers
99 views

Free finitely generated 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 ...
3
votes
2answers
225 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 ...
1
vote
1answer
134 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 ...
1
vote
2answers
117 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 ...
23
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 ...
9
votes
2answers
678 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 ...
16
votes
4answers
1k 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 ...
7
votes
3answers
967 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 ...
8
votes
4answers
268 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 ...
10
votes
1answer
711 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
194 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 ...
5
votes
6answers
1k 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 ...
5
votes
1answer
224 views

Existence proof of the tensor product using the Adjoint functor theorem.

Can one prove the existence of the tensor product by the adjoint functor theorem? (of, say, modules over a commutative ring) If yes, how would one check the SSC (solution set condition) for the hom ...
3
votes
1answer
219 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 ...
4
votes
4answers
662 views

Noetherian module implies Noetherian ring?

I know that a finitely generated $R$-module $M$ over a Noetherian ring $R$ is Noetherian. I wonder about the converse? I believe it to be false and I am looking for counterexamples. Also I wonder if ...
3
votes
2answers
320 views

Characterization of short exact sequences

The following is the first part of Proposition 2.9 in "Introduction to Commutative Algebra" by Atiyah & Macdonald. Let $A$ be a commutative ring with $1$. Let $$M' ...
2
votes
1answer
262 views

Divisible module which is not injective

On searching for some example of divisible module but not injective, I come across one in T.Y.Lam, Lectures on Modules and Rings. He considers the $\mathbb{Z}[x]$-module ...
1
vote
4answers
189 views

Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$? [closed]

Here $A$, $B$ and $C$ are $R$-modules. Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$? And what if $B$ and $C$ are finitely generated?
1
vote
1answer
123 views

Property of an operator in a finite-dimensional vector space $V$ over $R$

Let $L: V\to V$ be an operator in a finite-dimensional vector space $V$ over $R$. For any $n \geq 0$, let $K_n = \ker (L^n)$, $I_n = \mathrm{Im}(L^n)$. (a) Prove that there exists $N$ such that ...
12
votes
3answers
961 views

Homomorphisms of graded modules

Let $M$ and $N$ be graded $R$-modules (with $R$ a graded ring). $\varphi:M\rightarrow N$ is a homogeneous homomorphism of degree $i$ if $\varphi(M_n)\subset N_{n+i}$. Denote by $\mathrm{Hom}_i(M,N)$ ...
5
votes
1answer
271 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 ...
5
votes
1answer
483 views

Why are these all the indecomposable projective modules?

Let $A$ be a finite dimensional associative algebra with unit over a commutative field $k$. Suppose that $M$ is a finitely generated left module. Denote by $I(M)$ the injective hull of $M$ and by ...
5
votes
1answer
681 views

Flat not projective, projective not free [duplicate]

I am looking for examples of a flat but not projective module, and of a projective but not free module.
5
votes
1answer
224 views

restriction of scalars, reference or suggestion for proof

Let $f:R \rightarrow R'$ be a ring homomorphism that is epic in the category of rings. Let $M,N$ be $R'$-modules. Why is it that a homorphism $h:M \rightarrow N$ is $R'$-linear if and only ...
4
votes
2answers
184 views

What exactly is a trivial module?

Yes, this is a quite basic answer, but I have to admit to be absolutely confused about this notion. Searching on the web, I managed to found two possible definition of trivial modules, referring ...
3
votes
1answer
255 views

Euler's theorem: [3]^2014^2014 mod 98

Calculate without a calculator: $$\left [ 3 \right ]^{2014^{2014}}\mod 98$$ I know I have to use Euler's Theorem. As a hint it says I might need to use the Chinese Remainder theorem too. I know ...
3
votes
1answer
134 views

Is a stably free module always free? [duplicate]

Let $R$ be a commutative ring with unity and $M$ an $R$-module. If $M\oplus R^m\cong R^n$ for some integers $m,n \geq 1$ then must $M$ be finitely generated and free? Can somebody help me with ...
1
vote
2answers
85 views

An ideal which is not finitely generated

Let $K$ be a field and $A=K[x_1,x_2,x_3,...]$. Prove that the ideal $I:=\langle x_i: i \in \mathbb N\rangle$ is not finitely generated as $A$-module. I have no idea what can I do here, I mean, ...
1
vote
1answer
341 views

On the grade of an ideal

I need to prove the following statment (actually a special case of it). Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $I$ an ideal of $R$. Then $\operatorname{grade}(I,M)\geq 2$ if ...
18
votes
4answers
651 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 ...
12
votes
4answers
2k views

Intuitive explanation of Nakayama's Lemma

Nakayama's lemma states that given a finitely generated $A$-module $M$, and $J(A)$ the Jacobson radical of $A$, with $I\subseteq J(A)$ some ideal, then if $IM=M$, we have $M=0$. I've read the proof, ...