4
votes
1answer
121 views

Free modules basic understanding problem

I have been told that the $\mathbb{Z}$ module $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$ isn't free. For a module to be free, there must exist a subset such that every element is expressible as a finite linear ...
1
vote
1answer
16 views

Irreducibles - difficulty with the definition

I'm working from the definition that in an integral domain $R$, an irreducible is an element $p$ such that if $p=xy$ then either $x$ or $y$ is a unit. In certain proofs on my course, the lecturer has ...
1
vote
1answer
28 views

Unqiue Factorization Domains, is the product finite?

Having looked around a bit, the most common definition of a UFD is an integral domain such that any element can be expressed as a product of a unit and irreducible elements, and that this ...
1
vote
1answer
51 views

Question concerning a list sorting problem

I have the following question: Let $a,b,c,d$ be four natural numbers with $a \leq b$ and $c\leq d$. I have written a program that produces a list, which has as entries all 2-tuples $(x,y)$ with ...
1
vote
0answers
50 views

Direct proof that $\mathbb{Z}/p\mathbb{Z}$ is not a flat $\mathbb{Z}/p^2\mathbb{Z}$-module.

I am trying to prove that $\mathbb{Z}/p\mathbb{Z}$ is not a flat $\mathbb{Z}/p\mathbb{Z}$-module. The reasoning I have is the following. We have an exact sequence $0 \to \mathbb{Z}/p\mathbb{Z} ...
0
votes
1answer
18 views

Finitely generated quotient module isomorphism

In an assignment, I was asked to prove that if $M$ is an $R$ module and $N\subseteq M$ is a submodule such that $M/N$ is finitely generated and free, then there exists a submodule $F$ of $M$ which is ...
1
vote
0answers
25 views

Example of a ring of finite global dimension with flat qu0tient

I've been thinking about this for quite a while but I cannot seem to find an example of If $k$ is a commutative ring of finite global dimension and I'm looking for a strictly not-commutative ...
1
vote
1answer
16 views

Finitely generated quotient module

In an assignment question, I'm a bit stumped by the following: Let $R$ be a commutative ring, $M$ be an $R$-module and $N$ an $R$-submodule of $M$. I was first asked to prove that if both $N$ and ...
2
votes
3answers
61 views

Contrasting definitions of bimodules? An illusion?

Recently my definition of a bimodule over a $k$-algebra has been challenged and I believe both definitions to be equivalent, am I wrong? Notation: $k$ is a commutative ring and $A$ is a (unital ...
0
votes
1answer
62 views

Rank of a module when the base ring is not a domain

Suppose $R$ is a commutative Noetherian local ring with $1$, which is not a domain. Let $M$ be a (non-free) finite $R$-module. What is meant by rank of $M$ in this case?
0
votes
2answers
71 views

Graded rings and modules - which book to refer to?

Could you recommend a book with a good treatment of the theory of graded rings and modules?
1
vote
2answers
36 views

An integral domain and its field of fractions.

I'm trying to solve the following problem: Let $R$ be a integral domain which is not a field and $K$ its fractions field. Show that a non-zero module $R$-homomorphism from $K$ to $R$ does not ...
0
votes
1answer
33 views

Characterization of right noetherian rings

Here's a quick question on noetherian rings. I know that for a ring $R$, the following are equivalent. $R$ is left noetherian Every finitely generated left $R$-module is noetherian Every submodule ...
2
votes
1answer
37 views

A two sided ideal of a Noetherian ring

$R$ is a left Noetherian ring with a minimal left ideal. Consider the set of minimal left ideals of $R$ ordered by inclusion. Then there is a maximal element $\mathfrak b= \bigoplus_{i\in I} \mathfrak ...
2
votes
3answers
46 views

Example of ring $R$ with ideals $I\neq J$ such that $R/I \cong R/J$ as modules

It's easy to prove that if $I$, $J$ are two-sided ideals and $R/I\cong R/J$ as modules over $R$, then $I=J$. What about left ideals? Is there a simple counterexample? I believe I've found an answer, ...
1
vote
1answer
21 views

ideal,ring,flat module,modules over R

Is there a characterization of modules (AND equivalent characterizations of rings R) over integral domains R with the property that each left ideal in R is flat?When all left ideals are ...
1
vote
0answers
41 views

Simple Cogenerator for the category of left $R$-modules

I want to prove that if the category of left $R$-modules has a simple cogenerator $M$, then $R$ is a simple ring. (Here, $R$ is an arbitrary ring with $1$.) My try is to take an ideal $I$ of $R$ ...
0
votes
2answers
42 views

$R^{(I)} \cong K \oplus H$ where $R^{(I)}$ is free but $K$ is not free

Let $R$ be a commutative ring with unit. Is there an example of a direct sum of $R$-modules $$R^{(I)} \cong K \oplus H$$ where $R^{(I)}$ is free but $K$ is not free ? Clearly $R$ can't be a PID.
0
votes
0answers
21 views

Classification of separable algebras over a commutative ring

A separable algebra over a field can be classified as a finite product of matrix algebras over division algebras whose centers are finite dimensional separable field extensions of the field. (See ...
2
votes
1answer
42 views

Kahler forms of a smooth affine algebra vanish eventually?

If $k$ is a Noetherian ring, then do the Kahler forms of a smooth affine $k$-algebra of dimension $d$ vanish above $d$? I mean is: $\Omega^{d+1}_{A|k}\cong 0$?
1
vote
0answers
82 views

What is $\operatorname{Ass}\operatorname{Ext}^i(M,N)$?

This is exercise 1.2.27 of Bruns-Herzog: Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $N$ an arbitrary $R$-module. Deduce that $\operatorname{Ass}(\operatorname{Hom}_R(M,N)) = ...
2
votes
0answers
71 views

Automorphism of certain f.g. free modules

This is a quick question from Frohlich and Taylor's Algebraic Number Theory, II.4, p 94. Let $R$ be a Dedekind domain with quotient field $K$, $\mathfrak p$ is a non-zero prime ideal of $R$ and ...
1
vote
1answer
32 views

Basis for the completion of a free module

This (or similar) question might have been asked before- apologies for any duplication. I've got a Dedekind domain $R$, a non-zero prime ideal $P$ of $R$ and the completion $\widehat{R}$ of $R$ wrt ...
2
votes
1answer
45 views

How can I prove commutative rings are stably finite?

I have a proof that proves that a stably finite ring (i.e. one where $AB=I\iff BA=I$ for any two square matrices $A,B$ with entries in the ring) has the Invariant Basis Number (IBN). Trouble is, it is ...
0
votes
1answer
45 views

tensor product of R-algebra and f.g module [closed]

$R$ is a commutative noetherian ring. If $S$ is an $R$-algebra, and $M$ a finitely generated $R$-module, is $M\otimes_RS$ finitely generated $S$-module? I only need a hint. Thanks!
2
votes
1answer
51 views

Maximal linearly independent sets in a f.g. module

Suppose $M$ is a finitely generated module over a commutative unital ring $R$. Is it true that every maximal linearly independent set in $M$ has the same size? What is the most general condition ...
5
votes
1answer
69 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 ...
2
votes
1answer
20 views

Show that if $R$ is principal, $N $ is pure and $Ann(x+N)= Rd$ then there exists $y \in M$ such that $x+N=y+N$ and $Ann(y)=Rd$

Let $R$ a integral domain and $M$ a $R$-module. A submodule $N$ of $M$ is pure if for all $x \in M$ and $a \in R$ such that $ax \in N$, there exists $y \in N$ such that $ax=ay$. Show that if $R$ ...
1
vote
1answer
45 views

Basis of the localization of a free module is the localization of its basis?

Let $R$ be a ring and $M$ a free $R$-module with basis $B$. If $P$ is a prime ideal of $R$, is $B_P$ a basis for the $R_P$-module $M_P$?
1
vote
1answer
66 views

Show that $\operatorname{End}_{\mathbb{Z}}(\mathbb{Q})$ is isomorphic to the field $\mathbb{Q}$

I have problem in showing that $\operatorname{End}_{\mathbb{Z}}(\mathbb{Q})$ is isomorphic as a ring to the field $\mathbb{Q}$. Any idea? Thanks
1
vote
0answers
20 views

Monotonic Union of Submodule

The theorem is: Suppose $M$ is an $R$-module, $T$ is an index set, and for each $t \in T$, $N_t$ is a submodule of $M$. If $\{N_t\}$ is a monotonic collection, $\cup_{t\in T} N_t$ is a submodule. Is ...
5
votes
1answer
91 views

On the indecomposable decomposition of the reduction of an integral representation

Here is a problem I have been grappling with all day. I started out thinking it might be true but am now inclined to believe it is false, and would like to see a counterexample. Suppose $G$ is a ...
2
votes
2answers
124 views

Applications of groups, rings and modules in life [closed]

I have read calculus. Now I'm reading algebra. What is the application of groups, rings and modules in life? When we study derivations we know several applications. What about groups, rings and ...
1
vote
1answer
25 views

Show that if $M$ is a R-module free, for all $r \in R$ and $m \in M$ such that $r \neq 0$ e $m \neq 0$ we have $rm \neq 0$.

is it true that if $M$ is a $R$-module free, for all $r \in R$ and $m \in M$ such that $r \neq 0$ e $m \neq 0$ we have $rm \neq 0$. Comments: I tried to do the following: Suppose that $rm = ...
3
votes
1answer
27 views

Minimal set of generators VS Length of a module.

Recently, I have been thinking about a problem in which I try to use the information I have about the set of generators of a finitely generated module over a nice ring and say something about the ...
2
votes
1answer
74 views

Jacobson radical of a ring finitely generated over $\mathbb Z$

If a commutative ring $R$ with $1$ is finitely generated over $\mathbb Z$ could one deduce that the Jacobson radical of $R$ is nilpotent? I am aware of the well-known fact that when $R$ is ...
1
vote
0answers
28 views

An inverse limit of a certain inverse system

Let $∆$ be a directed set and $(N_i,f_{ji})_{i∈∆}$ be an inverse system of $R$-modules. Fix $α \in∆$ and consider $(M_i,g_{ji})_{i\in∆}$ as follows: $M_i=N_i$ for $i≥α$, $M_i=0$ for $i<α$, and ...
2
votes
1answer
72 views

Integral domain with a finitely generated non-zero injective module is a field

Suppose that $R$ is a integral domain. Suppose that there exists a non-zero finitely generated injective module $M$. How can I prove that $R$ is field?
3
votes
1answer
61 views

Direct product of Cohen-Macaulay rings/Eisenbud, Exercise 18.6

Somehow I believe (or doubt (!)) that direct product of two Cohen-Macaulay (C-M) rings may not be C-M. Can anybody give me an example verifying this? I would be grateful to him/her.
0
votes
0answers
56 views

Sufficient conditions for quotient ring to be Cohen-Macaulay

We know that every Noetherian integral domain with (Krull) dimension $1$ is Cohen-Macaulay (CM). In a commutative algebra text the author have presented the following problem: "Let $(R,m)$ be a CM ...
2
votes
1answer
28 views

Difference of a ring and its module over itself

What is the difference of a ring $R$ and the module $R_R$? It looks like they are just the same thing. Is it correct that the difference is $R$ is a ring with multiplication defined but $R_R$ is an ...
1
vote
1answer
25 views

Morita equivalence: Is $_{\mathrm{End}_R(P)}P$ projective if $P_R$ is?

Assume $P$ is a right projective $R$-module. Is $P$, viewed as a left $\mathrm{End}_R(P)$-module, projective as well? If not, under what conditions does it hold? Context: I am trying to ...
2
votes
1answer
27 views

Irredundant intersection of submodules

Let $A$ be a commutative ring, $M$ an $A$-module, and $N_\alpha\subset M$ a family of submodules. Consider the intersection $$\bigcap_\alpha N_\alpha.$$ We say that the intersection is irredundant if ...
2
votes
1answer
82 views

Quasi-hereditary algebras

Can anyone please recommend a reference (I prefer a book chapter) on Quasi-hereditary algebras? and if it is possible tell me how to prove that Schur algebras are Quasi-hereditary (or any other ...
0
votes
1answer
30 views

$\hom_R(A,B)$ is finitely generated if $R$ is noetherian [duplicate]

This is part of an exercise I'm doing, from Rotman Introduction to homological algebra. Let $R$ be a commutative ring, and let $A$ and $B$ be finitely generated $R$-modules. Then if $R$ is ...
2
votes
1answer
45 views

Starting projective modules problems

Show that if $n=rs$ where $n,r,s>1$ are positive integers, then the $ \mathbb{Z}_n $-module $r \mathbb{Z}_n$ is projective but it is not free if $(r,s)=1$. Any ideas or help how to prove this ...
2
votes
1answer
72 views

Newbie into categorical proofs

Let $$ F:Mod_A \to Mod_B $$ an aditive , exact and covariant functor and $$ M ∈ Mod_A $$ and $$M_1 , M_2 $$ submodules of M . Show that $$ F( M_1\cap M_2)=F(M_1)\cap F(M_2 ) $$ $$ F( M_1+ ...
0
votes
2answers
37 views

Not so usual equivalence of maximal left ideal of a ring

I was reading Foundations of Module and Ring Theory and i found this equivalence of maximal left ideal as exercise in the the first chapter: A left ideal $I$ of a ring $R$ is a maximal if and only ...
-1
votes
1answer
27 views

Chains in modules

I'm answering this question: Let $M$ be a module. Show that if $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated but $A$ is finitely generated whenever ...
2
votes
1answer
36 views

$\bf{Z}$-module homomorphism and $\bf{Q}$-module homorphism

$R$-modules are also $\bf{Z}$-modules and $R$-module homomorphisms are also $\bf{Z}$-module homomorphisms. If $M$ and $N$ are $\bf{Q}$-modules and $f : M \rightarrow N$ is a $\bf{Z}$-module ...