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

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19
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 ...
15
votes
1answer
551 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 ...
3
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 ...
6
votes
1answer
590 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 ...
13
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. ...
8
votes
1answer
883 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 ...
8
votes
1answer
651 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
864 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 ...
4
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 ...
11
votes
4answers
5k 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 ...
16
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 ...
5
votes
2answers
112 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 ...
3
votes
1answer
63 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 ...
3
votes
2answers
151 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
2answers
105 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
124 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 ...
21
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
542 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
854 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
1answer
571 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 ...
7
votes
4answers
246 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 ...
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 ...
3
votes
4answers
540 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
1answer
207 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
120 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
386 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
234 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
385 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
205 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 ...
2
votes
1answer
122 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 ...
1
vote
1answer
300 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 ...
14
votes
3answers
548 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 ...
14
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 ...
8
votes
1answer
1k views

Why do direct limits preserve exactness?

I've heard that taking direct limits is an exact functor in the category of modules, and I'm trying to figure out why, as I couldn't find a proof. Suppose you have homomorphisms $\varphi_i: K_i\to ...
6
votes
2answers
205 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 ...
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
2answers
311 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 ...
8
votes
2answers
864 views

Direct summand of a free module

Let $M$, $L$, $N$ be $A$-modules and $M=N\oplus L$. If $M$ and $N$ are free, is $L$ necessarily free?
6
votes
1answer
843 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 ...
6
votes
2answers
321 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 ...
12
votes
1answer
3k 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 ...
10
votes
2answers
437 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 ...
7
votes
2answers
173 views

On a commutative diagram

Let a commutative diagram be given: $$\require{AMScd} \begin{CD} 0 @>>> A @>f>> B @>g>> C @>>> 0 \\ @. @V{\alpha}VV @V{\beta}VV @V{\gamma}VV @. \\ 0 @>>> ...
6
votes
1answer
84 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
2answers
90 views

Why does $M\otimes k(\mathfrak{m})=M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m}$? (From Matsumura, proof of Theorem 4.8.)

Matsumura's Commutative Ring Theory, proof of Theorem 4.8, page 27, says: Let $A$ be a ring, $M$ a finite $A$-module, and $\mathfrak{m}$ a maximal ideal. If ...
4
votes
1answer
76 views

$2 \otimes_{R} 2 + x \otimes_{R} x$ is not a simple tensor in $I \otimes_R I$

This question is related to Properties of the element $2 \otimes_{R} x - x \otimes_{R} 2$. Another exercise of Dummit-Foote is to show that Let $I = (2, x)$ be the ideal generated by $2$ and $x$ ...
4
votes
4answers
87 views

Proving that P/PJ is a projective right module over R/J

If P is a projective right module over a ring R and J is a two sided ideal of R. Prove that P/PJ is a projective right module over R/J . My idea was trying to proof that " $M$ is an $R$-module ...
2
votes
0answers
157 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
611 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 ...