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

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12
votes
2answers
1k 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. ...
3
votes
1answer
1k 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 ...
16
votes
5answers
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 ...
14
votes
1answer
475 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 ...
8
votes
1answer
841 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
1answer
523 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 ...
8
votes
1answer
802 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 ...
9
votes
4answers
4k views

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

Free modules are projective, and projective modules are direct summand of free modules. Is there any example of projective modules that are not free? (I know this is not possible for modules of ...
8
votes
1answer
515 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. ...
15
votes
5answers
2k 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 ...
8
votes
3answers
1k 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 ...
8
votes
2answers
459 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 ...
6
votes
2answers
792 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 ...
7
votes
4answers
228 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 ...
4
votes
6answers
887 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 ...
2
votes
1answer
194 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 ...
1
vote
1answer
118 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 ...
6
votes
3answers
334 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 ...
5
votes
1answer
212 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
339 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
194 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
3answers
836 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 ...
3
votes
4answers
463 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 ...
2
votes
2answers
97 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 ...
1
vote
1answer
287 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 ...
13
votes
3answers
498 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
893 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 ...
6
votes
2answers
195 views

An $R$ module and $S$ module that cannot be an $R$-$S$ bimodule

In connection with this question: Modules and tensor products Question: For two commutative rings $R$ and $S$ (with unity), is there an abelian group $M$ which has $R$ module and $S$ module ...
12
votes
2answers
304 views

What does $R-\mathrm{Mod}$ tell us about $R$?

I have read the phrase "a good way to study a ring R is to study its modules" many times - though I cannot really see why this should be a general phenomenon. Is this phenomenon an analogue to (or an ...
6
votes
2answers
318 views

Whence this generalization of linear (in)dependence?

I recently came across a definition of (in)dependence that is supposed to be a generalization of linear (in)dependence among a set of vectors: An element $x$ is dependent on a set of elements ...
14
votes
2answers
1k views

Short Exact Sequences & Rank Nullity

This is a well known lemma that consistently appears in textbooks, either as a statement without proof, or as an exercise (see for example pp. 146 of Hatcher) If $0 \stackrel{id}{\to} A ...
11
votes
1answer
2k views

Finitely generated module with a submodule that's not finitely generated

Can someone give an example of a ring $R$, a left $R$-module $M$ and a submodule $N$ of $M$ s.t. $M$ is finitely generated, but $N$ is not finitely generated ? I tried a couple of examples of ...
9
votes
2answers
329 views

Finitely generated flat modules that are not projective

Over left noetherian rings and over semiperfect rings, every finitely generated flat module is projective. What are some examples of finitely generated flat modules that are not projective? Compare ...
6
votes
1answer
81 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 ...
4
votes
1answer
685 views

Specific proof that any finitely generated $R$-module over a Noetherian ring is Noetherian.

I have seen a handful of proofs that any finitely generated module over a Noetherian ring is again Noetherian. I'm specifically trying to understand the following proof idea. It goes as this: Observe ...
2
votes
0answers
149 views

Associated prime ideals of Hom (Bruns and Herzog, exercise 1.2.27)

Let $R$ be a Noetherian ring and $M,N$ finitely generated modules. I want to show that $$\mathrm{Ass}_R(\mathrm{Hom}_R(M,N)) = \mathrm{Ass}_R(N) \cap \mathrm{Supp}(M).$$ I don't understand what ...
10
votes
3answers
558 views

Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant?

Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$? This math.SE question deals with the case that ...
6
votes
2answers
1k 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 ...
5
votes
1answer
171 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 ...
4
votes
2answers
320 views

Showing $R(m\otimes n)$ is free if $Rm$ and $Rn$ are free

Let $R$ be a commutative ring with identity. The following is a statement I came across about the submodule $Rt$ generated by a decomposable tensor $t=m\otimes n$ being free, given that $Rm$ and ...
2
votes
2answers
244 views

Proof of $M$ Noetherian if and only if all submodules are finitely generated

Is my proof of proposition 6.2 on page 75 correct? (it's different from what they do in the book) Proposition 6.2.: $M$ is a Noetherian $A$ module $\iff$ every submodule of $M$ is finitely generated. ...
1
vote
2answers
63 views

About definition of “direct sum of $p$-vector subspaces”

In the books 1 and 2, in Somme directe d'une famille de sous-espaces vectoriels, I am reading the following: 1) let $E,F$ two vector subspaces of $V$, $E+F$ is direct sum, $E+F \doteq E\oplus F$, if ...
1
vote
1answer
2k views

The (Jacobson) radical of modules over commutative rings

Let $M$ be a module over a commutative ring $R$. Let $\Omega$ be the set of all maximal ideals of $R$. Prove that $\operatorname{Rad}(M)=\bigcap_{\mathfrak m\in \Omega}\mathfrak mM$, where ...
1
vote
2answers
529 views

Finitely generated module 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.
11
votes
2answers
711 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
174 views

Modules which are isomorphic to their tensor product.

Suppose that we have a commutative ring $R$. I am interested in finding the (finitely generated and projective, if you want) $R$-modules $M,$ such that $M\cong M\otimes_R M$ as $R$-modules. I know ...
5
votes
1answer
158 views

Does similarity of integer matrices imply the transition matrix is an integer matrix?

I'm working on a homework question, and I'm stuck. The question is: Let $A$ and $B$ be $2n \times 2n$ rational matrices with $A^2=B^2=-Id$. The first part of the question asks to show that $A$ and ...
5
votes
2answers
151 views

Associated primes of a sum of modules

Let $M$ be a module with $M_1$ and $M_2$ submodules such that their sum (not necessarily a direct sum) is $M$. Is it true in full generality that $\text{Ass}(M) = \text{Ass}(M_1) \cup ...
4
votes
1answer
531 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.
3
votes
1answer
84 views

Integral domain with a finitely generated non-zero injective module is a field

Suppose that $R$ is a integral domain. Suppose that there exists a non-zero finitely generated injective module $M$. How can I prove that $R$ is field?