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

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34
votes
3answers
3k 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 ...
21
votes
3answers
384 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 ...
21
votes
3answers
366 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, ...
19
votes
2answers
1k 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 ...
18
votes
4answers
847 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 ...
18
votes
3answers
298 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 ...
16
votes
4answers
379 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 ...
16
votes
1answer
186 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
415 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 ...
14
votes
4answers
1k 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
4answers
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 ...
14
votes
2answers
145 views

If $M\otimes N=R^n$ need $M$ be projective?

So if over a commutive ring, $R$, we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ be finitely generated projective? We have finite generation, because if $M\otimes N$ is ...
14
votes
1answer
399 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 ...
13
votes
3answers
415 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 ...
13
votes
2answers
2k 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 ...
13
votes
2answers
936 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 $\operatorname{Ann}(M)+\operatorname{Ann}(N)=A$. ...
12
votes
2answers
901 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
2answers
296 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
2answers
472 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.
12
votes
2answers
145 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 ...
12
votes
1answer
275 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 ...
11
votes
2answers
990 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
130 views

How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module?

How many ways are there to consider $\Bbb Q$ as an $\Bbb R$-module? I guess there is only one way, and that is the trivial case. i.e. $$\forall r\in \Bbb R,\, \forall a,b\in \Bbb Q \qquad r\cdot ...
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 ...
11
votes
2answers
541 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 ...
11
votes
3answers
183 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 ...
11
votes
1answer
444 views

Why do we call it trace?

For any module $P$, we define $\mathrm{tr}(P)=\sum \mathrm{im}(f)$, where $f$ ranges over all elements of $\mathrm{Hom}(P,R)$, and call it trace. Why does it have such a name? Does it have any ...
11
votes
3answers
628 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 $\mathrm{Hom}_i(M,N)$ ...
10
votes
4answers
751 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
7answers
880 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 ...
10
votes
4answers
105 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 $Hom_R(M,N)$ has the structure of an $R$-module? I was reading the preface of the Homological Algebra book by Rotman and ...
10
votes
2answers
475 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 ...
10
votes
1answer
328 views

Finitely generated modules over PID

Let $A$, $B$, $C$, and $D$ be finitely generated modules over a PID $R$ such that $A\oplus $ $B$ $\cong$ $C\oplus $ $D$ and $A\oplus $ $D$ $\cong$ $C\oplus $ $B$ . Prove that $A$ $\cong$ $C$ and $B$ ...
10
votes
3answers
78 views

An example of a noncommutative PID

It's well known that when a ring $R$ is a PID, every submodule of a free $R$-module is free. I'm interested in cases when the converse holds -- that is, in rings $R$ which have the property that every ...
10
votes
1answer
107 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 ...
10
votes
2answers
303 views

Ring homomorphisms which yield same endomorphisms on modules

Let $\varphi: R \rightarrow S$ be a (unital) ring homomorphism. So every left $S$-module $M$ has also a left $R$-module structure via $\varphi$ and in general we have $$ \text{End}_S(M) \subseteq ...
9
votes
4answers
238 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} ...
9
votes
2answers
550 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 ...
9
votes
3answers
484 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 ...
9
votes
1answer
123 views

diagonalizing a matrix over the $\ell$-adics

Let $M$ be a $2 \times 2$ matrix with coefficients in $\mathbb{Z}_{\ell}$ whose characteristical polynomial is $$ P(T) = T^2- (a+d) T + (ad-bc). $$ I've encountered the following assertion: If ...
8
votes
4answers
178 views

On Similarities and Differences Between Right and Left Modules over a Ring

I have read that for a ring $ R $ in general, right $ R $-modules are not the same things as left $ R $-modules. Why do we say that a right $ R $-module is equivalent to a left $ R $-module only ...
8
votes
4answers
632 views

Remainder when $20^{15} + 16^{18}$ is divided by 17

What is the reminder, when $20^{15} + 16^{18}$ is divided by 17. I'm asking the similar question because I have little confusions in MOD. If you use mod then please elaborate that for beginner. ...
8
votes
1answer
396 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
2answers
396 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 ...
8
votes
1answer
731 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
83 views

Why over $\mathbb{Z}/n\mathbb{Z}$ projectivity, injectivity and flatness coincide for cyclic modules?

Assume $R=\mathbb{Z}/n\mathbb{Z}$ ($n\neq0$) and let $M$ be a cyclic $R$-module. Could you tell me how to prove that $M$ is projective if and only if it is injective if and only if it is flat? And ...
8
votes
2answers
64 views

Zorn's Lemma in noetherian modules

For noetherian modules, we have in particular the equivalent definitions that the Ascending Chain Condition holds and that every nonempty subset of submodules has a maximal element. Now I can prove ...
8
votes
2answers
255 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 ...
8
votes
2answers
102 views

$R$ with an upper bound for degrees of irreducibles in $R[x]$

One very convenient property of $\mathbb{R}$ as a ring is that there is an upper bound for the degree of irreducible polynomials in $\mathbb{R}[x]$, as If $f\in\mathbb{R}[x]$ has degree larger ...
8
votes
1answer
386 views

is the pushforward of a flat sheaf flat?

Let $f:X \to Y$ be a morphism of schemes and let $F$ be an $\mathcal{O}_X$-module flat over $Y$. Is $f_*F$ flat over $Y$? What's wrong with this argument? [EDIT: as Parsa points out, the (underived) ...