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

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57
votes
3answers
6k views

What is the Tor functor?

I'm doing the exercises in "Introduction to commutive algebra" by Atiyah&MacDonald. In chapter two, exercises 24-26 assume knowledge of the Tor functor. I have tried Googling the term, but I don'...
27
votes
5answers
5k 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 ...
26
votes
2answers
3k views

Tensor products commute with direct limits

This is Exercise 2.20 in Atiyah-Macdonald. How can we prove that $\varinjlim (M_i \otimes N) \cong (\varinjlim M_i) \otimes N$ ? Atiyah-Macdonald give a suggestion, they say that one should ...
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
762 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 ...
25
votes
1answer
765 views

Does Nakayama Lemma imply Cayley-Hamilton Theorem?

Consider the Cayley-Hamilton Theorem in the following form: CH: Let $A$ be a commutative ring, $\mathfrak{a}$ an ideal of $A$, $M$ a finitely generated $A$-module, $\phi$ an $A$-module endomorphism ...
24
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 the ...
22
votes
3answers
484 views

My proof of $I \otimes N \cong IN$ is clearly wrong, but where have I gone wrong?

Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes: Let $R$ be a ...
22
votes
3answers
472 views

$\operatorname{Ann}(M\otimes_A N)=\operatorname{Ann}M+\operatorname{Ann}N$?

In the course of working on an exercise in Atiyah-MacDonald (exercise 3 on p. 31), I've come to the belief that, for $A$ an arbitrary commutative ring and $M,N$ arbitrary $A$-modules, $$\operatorname{...
22
votes
1answer
5k views

Finitely generated module with a submodule that is not finitely generated [duplicate]

Can someone give an example of a ring $R$, a left $R$-module $M$ and a submodule $N$ of $M$ such that $M$ is finitely generated, but $N$ is not finitely generated? I tried a couple of examples of ...
21
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 ...
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 finitely-...
20
votes
2answers
2k views

Applications of the Jordan-Hölder Theorem.

All books about group theory bring the Jordan-Hölder theorem, but does not give applications. I only saw in Rotman's book the Fundamental Theorem of Arithmetic being proved as a consequence of this ...
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 ...
19
votes
2answers
490 views

If the tensor power $M^{\otimes n} = 0$, is it possible that $M^{\otimes n-1}$ is nonzero?

Let $M$ be a module over a commutative ring $R$. It is possible that $M \otimes M = 0$ if $M$ is nonzero, for example when $R = \mathbb{Z}$ and $M = \mathbb{Q}/ \mathbb{Z}$. What about when higher ...
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 ...
19
votes
3answers
381 views

Bound on nilpotency index of endomorphisms

Let $A$ be a Noetherian ring (commutative with $1$) and $M$ a finitely generated $A$-module. I want to show that there exists a bound $n$ such that for every nilpotent endomorphism $T : M \to M$ we ...
18
votes
1answer
2k 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 ...
18
votes
1answer
481 views

Modules with $m \otimes n = n \otimes m$

Let $R$ be a commutative ring. Which $R$-modules $M$ have the property that the symmetry map $$M \otimes_R M \to M \otimes_R M, ~m \otimes n \mapsto n \otimes m$$ equals the identity? In other words,...
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. ...
17
votes
4answers
462 views

Non-isomorphic exact sequences with isomorphic terms

I love it when an undergraduate catches me out. I'm lecturing a first course in (not necessarily commutative) rings (with 1) and I've spent the last few weeks doing basic module theory. I defined a ...
17
votes
2answers
2k views

What does a zero tensor product imply?

I'm trying to prove that for two finitely generated $A$-modules $M,N$ ($A$ being any cmmutative ring), the tensor product $M\otimes_A N$ is zero iff $\operatorname{Ann}(M)+\operatorname{Ann}(N)=A$. ...
16
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 ...
16
votes
2answers
2k 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 \stackrel{f}{\...
16
votes
2answers
290 views

If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective

If over a commutative ring $R$ we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ are finitely generated projective? We have finite generation, because if $M\otimes N$ is ...
16
votes
2answers
832 views

Motivation behind the definition of flat module

Can someone explain what is the motivation behind the definition of a flat module? I saw the definition but I don't really know why it is important to work with these structures.
16
votes
1answer
217 views

Can extending a finite ground field make modules isomorphic?

$\def\Hom{\mathrm{Hom}}$Let $k$ be a field, $A$ a $k$-algebra and let $M$ and $N$ be $A$-modules, finite dimensional over $k$. Let $K$ be an extension of $k$, so $A \otimes K$ is a $K$-algebra and $M \...
16
votes
1answer
746 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 ...
15
votes
2answers
103 views

What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$?

What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$? Also, since $\mathbb{Z}[i]$ is a PID, we should be able to write this $\mathbb{Z}[i]$-module as a direct sum of cyclic $\mathbb{Z}[i]$-...
15
votes
2answers
222 views

Questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$

Setup: Let $A$ be a self-injective algebra (so projective = injective for modules) and let $D^b(A)$ and $K^b(A)$ be the bounded derived category and the full subcategory consisting of the perfect ...
15
votes
1answer
272 views

What is the importance of modules in algebraic geometry?

I have been trying to teach myself the basics of algebraic geometry. I understand the basic premise, how we define geometry spaces (algebraic sets and schemes) in terms of commutative rings. And I ...
14
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, ...
14
votes
1answer
366 views

Lemma on infinitely generated projective modules

Is it true that every finitely generated submodule of a non-finitely generated projective over a (not necessarily commutative!) ring is contained in a proper summand? (Ideally there's a standard ...
14
votes
1answer
56 views

Regard naturally as modules.

Suppose that $I$ is a two-sided ideal in the ring $R$, and that $M$ is a module over the quotient ring $R/I$. Why can we naturally regard $M$ as a $R$-module that is annihilated by $I$? Conversely, ...
13
votes
4answers
328 views

Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.

Disclaimer: Though I have been re-reading my notes, and have scanned the relevant texts, my commutative algebra is quite rusty, so I may be overlooking something basic. I want to show $\mathbb{Q} \...
13
votes
1answer
4k 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
3answers
169 views

Embeddings $A → B → A$, but $A \not\cong B$?

Are there any nice examples of structures (groups, modules, rings, fields) $A$ and $B$ such that there are embeddings $A → B → A$ while $A \not\cong B$? I would especially like to see an example for ...
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. I ...
13
votes
2answers
999 views

Universal property of free module, “converse”

Let $F$ be a free $R$-module with a basis $B$. We know that $B$ satisfies the following property: For any $R$-module $M$ and any $g:B\rightarrow M$, there exists a unique $R$-map $\varphi:F\...
13
votes
3answers
228 views

If $M\oplus M$ is free, is $M$ free?

If $M$ is a module over a commutative ring $R$ with $1$, does $M\oplus M$ free, imply $M$ is free? I thought this should be true but I can't remember why, and I haven't managed to come up with a ...
13
votes
3answers
1k 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 $\mathrm{Hom}_i(M,N)$ ...
12
votes
7answers
3k views

Applications of the Isomorphism theorems

In my study of groups, rings, modules etc, I've seen the three isomorphism theorems stated and proved many times. I use the first one ( $G/\ker \phi \cong \operatorname{im} \phi$ ) very often, but I ...
12
votes
2answers
339 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 ...
12
votes
4answers
195 views

Why is $\operatorname{Hom}(M,N)$ not necessarily an $R$ module?

Let $R$ be a ring, and $M,N$ be left $R$-modules. Then is it not true that $\operatorname{Hom}_R(M,N)$ has the structure of an $R$-module? I was reading the preface of the Homological Algebra book ...
12
votes
3answers
858 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 $R$...
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 ...
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 ...
12
votes
2answers
847 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 ...
12
votes
0answers
246 views

Non-reflexive module isomorphic to its double dual

Could you give me an example of a non-reflexive module isomorphic to its double dual? I found an example here but I cannot understand it, do you have any simpler examples? By this question we ...
11
votes
2answers
730 views

Motivation behind the ingredients of First Cohomology group $H^1$

I started reading the Cohomology theory of groups. But I am not able to get any intuition or motivation behind the following : It is concerned with the formal definitions of crossed and principal ...