0
votes
0answers
19 views

Presentation of a module by generators and relations

Let $R:=\mathbb C[T]$. Match the $R$-module with the presentation by generators and relations. $\bullet$$R$-modules: $M:=\mathbb C[T,T^{-1}]$ (Laurent Polynomials)$\qquad$$N:=\mathbb ...
1
vote
2answers
26 views

An ideal which is not finitely generated

Let $K$ be a field and $A=K[x_1,x_2,x_3,...]$. Prove that the ideal $I:=\langle x_i: i \in \mathbb N\rangle$ is not finitely generated as $A$-module. I have no idea what can I do here, I mean, ...
1
vote
0answers
26 views

Locally cyclic module exercise

An $A$-module $M$ is locally cyclic if every submodule of $M$ of finite type (finitely generated) is cyclic. (i) Show that every submodule of a locally cyclic module is locally cyclic. (ii) Prove ...
1
vote
1answer
10 views

Module of finite type has a minimal system of generators

I am trying to show the following statement: Every module of finite type, i.e. finitely generated, has a minimal system of generators (a minimal system of generators $\mathcal S$ is a system of ...
1
vote
0answers
30 views

What is the tensor product of Z/10Z with Z/12Z? [duplicate]

I've just met tensor products of modules in part of my self-study and as a concrete exercise I've been asked to calculate $\mathbb{Z}/(10){\otimes}_{\mathbb{Z}} \mathbb{Z}/(12)$. Short of writing out ...
0
votes
2answers
20 views

Questions related to the concept of $k$-algebras

I am reading about modules and some days ago I've worked on some exercises related to $k$-algebras. The definition I've seen of $k$-algebra is that it is a field $k$ and a ring $A$ together with a ...
3
votes
1answer
33 views

Short exact sequence of modules

I am trying to show that if we have the following left splitting short exact sequence of $R-$modules: $0 \rightarrow M \stackrel{f} \rightarrow N \stackrel{g} \rightarrow S \rightarrow 0$ then there ...
1
vote
2answers
33 views

Why $\mathbb Z/ 2 \mathbb Z$ is not a free module?

I am reading some abstract algebra notes about free modules. It says that not all modules are free and the example to illustrate this is $\mathbb Z/ 2\mathbb Z$ (as a $\mathbb Z$-module) is not a free ...
5
votes
2answers
53 views

The rationals as a direct summand of the reals

The rationals $\mathbb{Q}$ are an abelian group under addition and thus can be viewed as a $\mathbb{Z}$-module. In particular they are an injective $\mathbb{Z}$-module. The wiki page on injective ...
0
votes
1answer
14 views

Morphism of $k$-algebras between abelian group of $n \times n$ matrices and $m \times m$ matrices

Problem Let $k$ be a field and $f:M_n(k) \to M_m(k)$ be a morphism of $k$-algebras ($n,m \in \mathbb N$). Prove that $n$ divides $m$. I have no idea what to do here, I thought that maybe I should ...
1
vote
1answer
29 views

Definition of Tensor Product of Modules

I am really struggling to understand several parts of the definition of tensor product given in my lecture notes: Definition of the tensor product *Denote by L the free A-module with a basis ...
1
vote
3answers
22 views

Definition of direct sum of modules?

I have just started studying modules, and am trying to figure out the definition of the direct sum of modules but I'm having trouble since different sources seem to give different definitions, for ...
2
votes
2answers
127 views

Vector space as simple $K[x]$-module

I am trying to solve the problem: Let $V$ be a vector space and $T$ a linear transformation $T:V \to V$. Let $(V,T)$ be a $K[x]$-module. Show that $(V,T)$ is simple if and only if $V$ is finite ...
2
votes
2answers
80 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 ...
0
votes
2answers
24 views

Linear independence and grammar

Let $A$ be a commutative ring. Normally, the linear independence is a property of a subset or a family elements of an $A$-module. But one often sees statements like "$v$ and $w$ are linearly ...
0
votes
2answers
41 views

$M$ is a simple module if and only if $M \cong R/I$ for some $I$ maximal ideal in $R$.

I am trying to show the following statement (taken from Rotman's Advanced Modern Algebra): Let $M$ be an $R$-module. Then $M$ is a simple module if and only if $M \cong R/I$ for some $I$ maximal ...
3
votes
1answer
49 views

Rank of projective module defined as the smallest $n$ such that $P$ is a direct summand of $R^n$

Over a commutative ring $R$, the rank of a projective module $P$ is defined by looking at the map $\text{rank}(P) : \text{Spec}(R) \rightarrow \mathbb{N}_0$ given by $\mathfrak{p}\mapsto ...
0
votes
0answers
19 views

A dual to essentially finitely generated notion

It is known that a (right) module $M$ over a ring $R$ has finite uniform dimension if and only if it is "essentially finitely generated", in the sense that there exists a finitely generated submodule ...
4
votes
0answers
59 views

Using the universal property of tensor product to show that $(\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}$ [duplicate]

In the algebra lecture i need to solve the following exercise Use the universal property of the tensor product to show that $$(\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong ...
0
votes
1answer
29 views

Problem related to modules and k-algebras

I am trying to do the following exercise: Prove that if $A$ is a $k-$algebra and $M$ is a module then the product $\lambda * m=\tau(\lambda)m$ ($\tau$ is the morphism from $K$ to $Z(A)$) defines on ...
0
votes
1answer
44 views

Lifting a direct summand of a free module

Suppose $R$ is a commutative ring, $I\subseteq R$ a principal ideal, and we're given split short exact sequences $ R \to R^n \to R^{n-1}$ and $ R/I \to (R/I)^n \to (R/I)^{n-1}$ the first inducing ...
0
votes
0answers
43 views

Augmentation ideal of tensor product of group rings

We have $\epsilon: \mathbb{Z}G\to\mathbb{Z}$ the augmentation such that $\epsilon(\sum z_gg)=\sum z_g$ and $\ker(\epsilon)=\operatorname{Aug}\mathbb{Z}G$. We have to $\mathbb{Z}(A\times A_1)\simeq ...
1
vote
0answers
29 views

Aluffi: submodule $\Longleftrightarrow$ cokernel?

Aluffi makes the following brief statement, in the context of modules: "The last sentence of Proposition 6.2 simply reiterates the slogan submodule $\Longleftrightarrow$ kernel and its mirror ...
8
votes
2answers
100 views

${\rm Hom}_R(M, R/M) =\{0\} \implies R$ is a field.

Let $R$ be a local ring with maximal ideal $M$. Suppose $M$ is finitely generated. Prove that if ${\rm Hom}_R(M, R/M) =\{0\}$, then $R$ is a field. ${\rm Hom}_R(M, R/M)$ stand for the group of ...
0
votes
0answers
21 views

quotients and direct sums

Let $H$, $K$, $W$, be submodules of a module $M$. Is it true that $(H \oplus K)/W \cong H/W \oplus K \cong H \oplus K/W$? The first seems to follow from 1st isomorphism theorem on the map $\phi = ...
1
vote
1answer
39 views

The condition that a ring is a principal ideal domain

If $R$ is a nonzero commutative ring with identity and every submodule of every free $R$-module is free, then $R$ is a principal ideal domain. What I don't know is how to show that every ideal is ...
1
vote
1answer
44 views

Does $ax\in\mathfrak{m}I$ with $x\in I\setminus\mathfrak{m}I$ and $a \in R$ imply $a\in\mathfrak{m}$ for an invertible fractional $R$-ideal $I$?

Let $R$ be an integral domain, $\mathfrak{m}$ a maximal ideal of $R$, and $I$ an invertible fractional $R$-ideal. If $x \in I \setminus \mathfrak{m}I$ and $a \not\in \mathfrak{m}$, do we have $ax ...
0
votes
1answer
29 views

Rank of abelian groups

I have read that given a $\mathbb{Z}$-module $M$, the maximal number of $\mathbb{Z}$-linear independent elements is given by $\operatorname{rank}M=\dim_\mathbb{Q}(\mathbb{Q}\otimes_\mathbb{Z}M)$. ...
-1
votes
0answers
36 views

Prove there are some elements in a commutative module [duplicate]

let R be a commutative ring and I is an ideal in R and also M is finite generating module on R. If $\varphi:\:M \to M$ be a homomorphism and $\varphi(M)\subset IM$ .Prove there are some elements ...
1
vote
1answer
29 views

Misunderstanding in Cartan-Eilenberg?

In Cartan Eilenberg's Homological algebra, page 13 it says: If $\Gamma$ is a principal ideal ring, then each ideal $I$ of $\Gamma$ is isomorphic with $\Gamma$, thus $I$ is free and $\Gamma$ is ...
1
vote
1answer
23 views

Projectiveness of direct product of projectives

It is well-known that a direct sum of modules is projective if and only if each summand is projective, and a direct product of modules is injective if and only if each of the modules are injective. ...
1
vote
0answers
49 views

Describe the automorphisms of this $\mathbf{Z}$-module

This follows up on what I thought was a good question which has now been deleted, asking about the automorphisms of the multiplicative group $\mathbf{Q}^{*}$. Is there a relatively simple ...
0
votes
1answer
29 views

counter example in $R$-modules law

Let $M$ be an $R$-module and $K$, $L$ and $N$ submodules. I would like to find a counterexample to the equality $$N\cap (K+L) =(N\cap K)+(N\cap L)$$ I can prove the equality is true when $K ...
3
votes
2answers
28 views

group of module homomorphisms is a module

I am trying to solve the following problem: Let $M$ and $N$ be two left $A$-modules. Prove that $Hom_A(M,N)$ has a left $Z(A)$-module structure with: $(a.f)(m)=a.f(m)$. Show $Hom_A(A,N) \cong N$ as ...
0
votes
0answers
24 views

A question about injection of a quotient in a direct summand

Let $M$ be a nonsingular right $R$-module in the sense that $Z(M)=\{m\in M : \operatorname{ann}(m)\text{ is essential in }R_R\}=0$. Could one find an $R$-monomorphism from the quotient $M/\bigcap N_i$ ...
0
votes
0answers
15 views

Goldie closure of the zero module

In "Lectures on Rings and Modules" by T.Y. Lam, it is claimed in Example (7.33) that the Goldie closure $cl(0)$ of the zero module $0$ on the ring $R=\left [\begin{array}\ \mathbb Z & \mathbb Z_2 ...
0
votes
1answer
82 views

exact sequence with infinite element

How to prove: Define exact sequence with infinite element from $R$-modules and $R$-homomorphism ,then exact sequence with infinite element from $\Bbb Z$-modules and $\Bbb Z$-homomorphism : ...
0
votes
1answer
64 views

Finite cyclic $\Bbb Z$-module and exact sequence

Suppose $M$ is $\Bbb Z$-module, cyclic, finite. How to prove $\require{AMScd}$ \begin{CD} 0 @>>> \Bbb Z @>>> \Bbb Z @>>> M @>>>0\\ \end{CD} is ...
-1
votes
1answer
46 views

a problem in exact sequence

Suppose $\require{AMScd}$ \begin{CD} @.\acute M @>f>> M @>g>>\check M @>>> O\\ @. @V \alpha V V\ @VV \beta V @VV \gamma V @. \\ O@>>> \acute N ...
1
vote
1answer
73 views

Finitely generated prime ideal and annihilator

Suppose $R$ is a commutative ring, $P$ is a prime ideal of $R$, $P$ is finitely generated, and $\operatorname{Ann}(P)=0$. Show that $$\operatorname{Ann}(P/P^2)=P.$$ These are my efforts: ...
0
votes
0answers
21 views

Module notation

Suppose I have a ring $R$ whose underlying group is $G$. Now suppose I read the definition: Let $M$ be an $n$-dimensional left module over $R$. Is this a conventional way of defining $M$ as ...
1
vote
1answer
46 views

A relation in a finitely generated module

Suppose $R$ is a commutative ring, $I$ is an ideal of $R$, and $M$ a finitely generated $R$-module s.t. $M=IM$. How to prove: $$\exists a \in I \text{ such that } (1-a)M=0. $$ I tried to solve: ...
1
vote
1answer
111 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 ...
7
votes
1answer
115 views

Tensor Product, Exterior Power and Splitting

Let $M$ be a $\mathbb{Z}$-module and consider the submodule $K=\langle m\otimes m\mid m\in M\rangle$ of $M\otimes M$. Under what conditions does the SES $$0\to K\to M\otimes M\to M\wedge M\to 0$$ ...
0
votes
1answer
28 views

Submodule iff subgroup?

It is late at night and time for another silly question: Is it true that a subset $S$ of an $R$-module $M$ is a submodule if and only if it is a subgroup of $M$ as an abelian group? Of course, by ...
0
votes
2answers
48 views

A question on endomorphisms of an injective module

This is a homework question I am to solve from TY Lam's book Lectures on Modules and Rings, Section 3, exercise 23. Let $I$ be an injective right $R$-module where $R$ is some ring. Let $H= ...
2
votes
1answer
32 views

Show the module of continuous functions of antiperiod $\pi$ is not cyclic.

Homework Hint Needed: Let $R$ be the ring of all continuous functions on $\mathbb{R}$ of period $\pi$. Let $S$ be the $R$-module of all continuous functions on $\mathbb{R}$ of antiperiod $\pi$. So ...
1
vote
1answer
43 views

Example of a commutative square without a map between antidiagonal objects?

In an abelian category, can there be a commutative diagram of (vertical/horizontal) exact sequences $$ \require{AMScd} \begin{CD} 0 @>>> N @>>> M\\ @. @VVV @VVV \\ 0 @>>> X ...
3
votes
1answer
42 views

Module over an associative ring with unity and axioms of projective geometry

According to Wikipedia(http://en.wikipedia.org/wiki/Projective_geometry#Axioms_of_projective_geometry), the axioms of projective geometry due to A. N. Whitehead are: G1: Every line contains at least ...
2
votes
0answers
32 views

Endomorphism rings of a $k$-algebra

Let $k$ be a field and $R$ be the $3$-dimensional $k$-algebra $\left [\begin{array}\ k & k \\ 0 & k \end{array} \right ]$. I have two conjectures: (1) Is any simple right $R$-module has ...