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

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3
votes
1answer
43 views

Why are Grothendieck's and Hartshorne's definitions of quasi-coherence equivalent?

Hartshorne's Algebraic Geometry defines an $\mathcal O_X$-module $\mathscr F$ to be quasi-coherent if there is an open affine cover $(U_i=\operatorname{Spec} A_i)_{i\in\mathcal I}$ of $X$ such that ...
1
vote
1answer
31 views

An odd expression appearing in proof that kernel and image of certain map on group ring $\mathbb Z[G]$ are equal

Let $G = \langle g \rangle$ be a cyclic group of order $n$. Consider the free ring $\mathbb Z[G]$ of all formal sums of elements from $G$ with coefficients from $\mathbb Z$, with multiplication given ...
5
votes
1answer
110 views

Show that $1\otimes (1,1,\ldots)\neq 0$ in $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n=2}^{\infty} (\mathbb{Z}/n \mathbb{Z})$.

Show that $1\otimes (1,1,\ldots)\neq 0$ in $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n=2}^{\infty} (\mathbb{Z}/n \mathbb{Z})$. Here's what I tried: If $1\otimes (1,1,\ldots)= 0$, then $1\otimes ...
1
vote
1answer
40 views

In what sense are complete local rings finitely generated modules?

In the first paragraph of section 18.4 of Eisenbud's Commutative Algebra, there is the following comment. Most interesting Noetherian rings can be written as finitely generated modules over ...
0
votes
2answers
50 views

Annihilator of extension of scalars vs. the extension the annihilatar

Let $A,B$ be commutative rings with 1, $f:A\to B$ a morphism of rings, $M$ an $A$-module, and $M_B=B\otimes_AM$ the extension of scalars. Then is it the case that $\text{Ann}(M)^e=\text{Ann}(M_B)$? ...
1
vote
1answer
15 views

ACC on chains of finitely generated submodules

Noetherian rings are those having ascending chain condition on ideals. There are also literatures concerning ACC on n-generated ideals (i.e., generated by n elements); see e.g, Commutative Rings with ...
0
votes
0answers
32 views

projective resolution of finitely generated modules

I am in the condition where I have a noetherian ring $R$ of finite global dimension. Consider the category of finitely generated (right) modules over $R$. Then I want to show that every module admits ...
16
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 ...
5
votes
1answer
76 views

Tensor product of $\mathbb{Q}$ with an infinite product [closed]

How can I prove that the tensor $\mathbb{Q} \otimes \left( \prod_n \mathbb{Z}/n\mathbb{Z} \right)$, where the product is taken over all the positive integers $n$, is not trivial?
6
votes
1answer
560 views

Examples proving that the tensor product does not commute with direct products [duplicate]

Examples proving why the tensor product does not distribute over direct products? In fact the canonical map is not surjective; can you give me a simple example?
2
votes
1answer
64 views

Homology commutes with direct product of chain complexes. Direct proof in a module category.

This is an attempt to prove that direct product of chain complexes commutes with homology (exercise in Weibel's book). I've had some success since I've proved that $Z_n(\prod_{\alpha \in A} C_{\alpha ...
3
votes
1answer
38 views

Verify proof that if $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian

I have to prove that If $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian We let $S$ be a non-finitely generated submodule of $M\otimes_R ...
0
votes
0answers
17 views

Torsion-free submodule of maximal rank

Let $A$ be a free $\mathbb{Q}[t_1^{\pm 1},\dots,t_\mu^{\pm 1}]$-module of rank $n$. If $M$ is a submodule of $A$, I know that it must be torsion free. Assuming the rank of $M$ is $n$ can I conclude ...
0
votes
0answers
30 views

General definition of free module over arbitrary rings and space of multi-linear functions

Consider definition of free module as given in this question: Free modules over commutative ring (possibly without unity) where free means having a LI spanning set I have proved if Can we say if V ...
3
votes
3answers
97 views

$\mathbb Z[1/2]$ is not finitely generated?

$\mathbb Z[1/2]$ is not finitely generated ? Maybe I misunderstood, what finitely generated means. Here it says, we need finitely many elements and I think $1$ and $1/2$ suffices as generators. ...
2
votes
0answers
31 views

Extending ring metric to arbitrary module

I am working on my own on an idea I had and this one is a bit difficult so I am seeking a bit of inspiration and to see what already exists. Let $R$ be a given ring with a metric $d$ and $M$ being an ...
3
votes
4answers
131 views

What is the difference between ring homomorphism and module homomorphism?

I know that $\mathbb{Z}$ and $2\mathbb{Z}$ are isomorphic as Z-modules. But they are not isomorphic as ring. I used to think that I can look at isomorphism as being "equal". However, now it seems that ...
2
votes
1answer
18 views

Z modules spanned by row space of matrix invariant under matrix multiplication

I have met this strange looking problem on which I have no idea, from my course on Abstract Algebra dealing with modules: Let $ v_1,...,v_k \in \mathbb{Z}^n $ row vectors of length n over $ ...
5
votes
1answer
150 views

Every finitely generated flat module over a ring with a finite number of minimal primes is projective

Over a commutative ring $R$, a finite type locally free (weak sense) module for which the rank function is locally constant is projective. If we notice that for each minimal prime $p$ of the ring, ...
2
votes
3answers
70 views

Is $A / \mathfrak{m}$ flat if $A$ is a local ring?

I'd like to prove the following: if $A$ is a local ring and $\mathfrak{m} \subset A$ its maximal ideal, then $A / \mathfrak{m}$ is a flat $A$-module. How can I do this? I've tried to find a suitable ...
7
votes
2answers
131 views

Problem with the ring $R=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q &\Bbb Q\end{bmatrix}$ and its ideal $D=\begin{bmatrix}0&0\\ \Bbb Q & \Bbb Q\end{bmatrix}$

Let us consider the ring $ R:=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q & \Bbb Q\end{bmatrix} $ and its two-sided ideal $ D:=\begin{bmatrix}0 & 0\\ \Bbb Q & \Bbb Q\end{bmatrix} $. Let then ...
3
votes
1answer
60 views

Ring of infinite global dimension which does not have a finitely generated module of infinite projective dimension

Let $R$ be a ring of infinite global dimension. A priori we can't immediately conclude that $R$ has a module of infinite projective dimension, since it could be the case that $R$ only has a sequence ...
1
vote
0answers
27 views

Constructing a ring consisting of formal infinite series from a given ring

Let $A$ be an $\mathbb{N}$-graded $\Bbbk$-algebra, where $\Bbbk$ is a field, and where $\dim_\Bbbk A_n < \infty$ for all $n \in \mathbb{N}$. I can't see anything preventing me from constructing a ...
0
votes
2answers
47 views

Prove that $\operatorname{Ext}^{d+1}(A, B)\cong \operatorname{Ext}^1(M_d,B)$

So, given a resolution, with $P_{i}$ projective modules: $$0\longrightarrow M_d\longrightarrow P_{d-1} \longrightarrow \cdots \longrightarrow P_0 \longrightarrow A\longrightarrow 0,$$ I'm trying to ...
2
votes
1answer
31 views

Kernel of a natural map is a direct summand of the covariant extension

I am reading chapter 2 of 'Homological Algebra' by Cartan and Eilenberg. 1/ Given a ring homomorphism $\varphi: \Lambda \rightarrow \Gamma$ and a right $\Gamma$-module $A$, we can treat $A$ and ...
3
votes
1answer
21 views

Is an injective map of $B$-modules also injective as an $A$-linear map if $B$ is an $A$-algebra?

I've been going through my submitted exercises again of my Commutative Algebra-class and I have the following question: Let $A$ be a commutative ring with unity. Given any injective homomorphism of ...
1
vote
2answers
279 views

Noetherian rings and modules

A ring is left-Noetherian if every left ideal of R is finitely generated. Suppose R is a left-Noetherian ring and M is a finitely generated R-module. Show that M is a Noetherian R-module. I'm ...
3
votes
2answers
995 views

Finitely generated modules 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.
0
votes
2answers
24 views

Quotient space decomposition

Let $V$ be a vector space(or module) with decomposition $V= V_1 \oplus V_2$. And let $W \subset V$ be a subspace with decomposition $W= W_1 \oplus W_2$ such that $W_1 \subset V_1$ and $W_2 \subset ...
0
votes
2answers
37 views

Checking commutativity of a diagram of modules over some ring and what the commutativity of the diagram implies.

Suppose that you have the following diagram of modules over some ring: These are my questions: (1) To prove that the diagram is commutative, we needs to prove that $gf=kh$, $wf=rv$, $zh=uv$, ...
0
votes
1answer
82 views

How do I prove (Module isomorphic direct sum of modules)

Module $M$ isomorphic to the external direct sum of modules $\bigoplus_{\alpha\in I}M_{\alpha}$ if and only if there are monomorphisms $i_{\alpha} : M_{\alpha} \rightarrow M, \alpha \in I$, such that ...
3
votes
1answer
49 views

Category of modules

For given a ring $R$, we can define the category of left $R$-modules. In fact, objects are all left $R$-modules and morphisms are $R$-module homomorphisms. NOw my question is: If we do not fix $R$, ...
1
vote
0answers
41 views

Free $R$-module when $R$ is not unital

We can easily construct free $R$-module when $R$ is unital by setting $$R[S] = \{ f\colon S\to R\,|\, f\ \text{finitely supported}\}$$ and defining operations pointwise. The key here is that we can ...
1
vote
2answers
65 views

Definition of exact sequence [closed]

Take a sequence $A \to B \to C \to 0$ of modules over a commutative ring. How would one show that it is exact? I understand the necessary surjectivity of $B \to C$, but what about the first map?
2
votes
1answer
41 views

Finite dimensional algebras with finite global dimension.

Let $A$ be a finite dimensional $k$-algebra, $k$ is a field, with a finite global dimension. I wonder if that implies $A$ is tame or finite type? or more generally is there a relation between these ...
0
votes
0answers
47 views

Comodule induced from a comodule over a graded coalgebra

Let $G$ be a group with $1$ be the identity element of $G$. Let $(C,\bigtriangleup,\epsilon)$ be a coalgebra over a field $K$. $C$ is called $G$-graded if $C$ admits a decomposition as a direct sum of ...
3
votes
2answers
51 views

Example of a $\mathbb{Z}$-module with exactly three proper submodules?

What is an example of a $\mathbb{Z}$-module which has exactly three proper submodules?
3
votes
3answers
666 views

Example of a Short Exact Sequence

I know that $A$ is a $\mathbb{Z}$-module. And I have a short exact sequence of the form $0 \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow A \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$ is ...
1
vote
1answer
40 views

$\mathcal{O}_L$ free over $\mathcal{O}_K[G]$

Let $L/K$ be a galois extension of number fields. Suppose $G:=\text{Gal}(L/K)$ is abelian. If $\mathcal{O}_L$ is free as $\mathcal{O}_K[G]$-module, is it true that it has rank 1?
-3
votes
1answer
54 views

$f:A\to B$ epimorphism if and only if $B/f(A)$ is a torsion module [closed]

I want to prove the double implication: $f: A\to B$ is an epimorphism in the category of torsion-free abelian groups $\Longleftrightarrow$ $B/f(A)$ is a torsion $\mathbb Z$-module. Thanks.
3
votes
3answers
67 views

Equivalent definition of R-module

I am studying a course of Commutative Algebra and I'm having trouble understanding an alternative definition of R-module. The standard definition is the following: Let $R$ be a ring (commutative, ...
2
votes
1answer
56 views

What is a periodic module

I've been reading the text "THE THEORY OF COMMUTATIVE FORMAL GROUPS OVER FIELDS OF FINITE CHARACTERISTIC" by Manin. On page 26, proposition 2.1, the author mentions the notion of a periodic module, ...
3
votes
0answers
38 views

Proving that Tor is a balanced functor using the derived category

At this end of this expository article on derived categories, R.P. Thomas says the following. There are two main advantages of this approach. Firstly that we have managed to make the complex ...
0
votes
0answers
28 views

What is the use of a right-module?

It seems that only the left-module provides a representation of a group. So what is the use of a right-module?
3
votes
1answer
27 views

How do you break up an exact sequence of any length to a “succession of short exact sequences”?

Note that if $\text{Hom}_R(D,-)$ functor takes short exact sequences to short exact sequences then it takes exact sequences of any length to exact sequences since any exact sequence can be broken ...
0
votes
1answer
36 views

Simple submodule of modules

I know that a simple right module is a non-zero right module $M_R$ whose submodules are only $M_R$ and $0$. Now fix a module $N_R$. If I want to show that a submodule $M_R$ of $N_R$ is simple what do ...
1
vote
1answer
42 views

Show that character vanishes on specific element $g$ if for $H \le G$ we have $[\chi_H, 1_H] = 0$ and all elements of $Hg$ are conjugate in $G$

Let $G$ be a finite group and $H \le G$ with $g \in G$ such that all elements of the coset $Hg$ are conjugate in $G$. Let $\chi$ be a $\mathbb C$-character of $G$ such that $[\chi_H, 1_H] = 0$. Show ...
1
vote
3answers
45 views

Doubts about adjunction of elements to a ring

I've just solved the following exercise Suppose we adjoin an element $\alpha$ to $\mathbb{R}$ satisfying the relation $\alpha^2 = 1$. Prove that the resulting ring is isomorphic to the product ...
-2
votes
1answer
23 views

Uniform modules and submodules [closed]

A non-zero left module $M$ over a ring with unity is called uniform if any two non-zero submodules have non-empty intersection; $M$ is said to contain enough uniforms if any non-zero submodule ...
2
votes
2answers
64 views

$\mathrm{Hom}_R(N,M)$ is an essential extension of $\mathrm{Hom}_R(N,L)$

$M$ is an $R$-module and an essential extension of $L$. $N$ is a finitely generated submodule of $M$. Then $\mathrm{Hom}_R(N,M)$ is an essential extension of $\mathrm{Hom}_R(N,L)$. I tried to use ...