Tagged Questions

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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 ...
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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 ...
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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 ...
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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 ...
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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, ...
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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 ...
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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 $R$ ...
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$R^{(I)} \cong K \oplus H$ 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.
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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 ...
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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: $\Omega^{d+1}_{A|k}\cong 0$?
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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 ...
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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 ...
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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 ...
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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?
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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.
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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 ...
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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 ...
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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 ...
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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 ...
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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 ...
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$\hom_R(A,B)$ 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 ...
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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 ...
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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+ ...
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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 ...
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 ...
$\bf{Z}$-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 ...