# Tagged Questions

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

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### Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?

Why isn't an infinite direct product of copies of $\Bbb Z$ a free module? Actually I was asked to show that it's not projective, but as $\Bbb{Z}$ is a PID, so it suffices to show it's not free. I ...
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### Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
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### 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 ...
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### 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 ...
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### Are all simple left modules over a simple left artinian ring isomorphic?

I've a basic question. $R$ is a simple left artinian ring. I want to show that all simple left $R$-modules are isomorphic. A simple $R$-module $M$ is isomorphic to $R/J$ where $J$ is a maximal ...
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### When $\operatorname{Hom}_{R}(M,N)$ is finitely generated as $\mathbb Z$-module or $R$-module?

Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is ...
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### for any ring $A$ the matrix ring $M_n(A)$ is simple if and only if $A$ is simple

Let integer $n\geq 1$. I have obtained that for any field $k$, the matrix ring $M_n(k)$ is simple, i.e., $M_n(k)$ contains no nonzero proper two sided ideals. Now I want to prove that: for any ring $A$...
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### 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. ...
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### Example of modules that are projective but not free; torsion-free but not free

Free modules are projective, and projective modules are direct summands of free modules. Are there examples of projective modules that are not free? (I know this is not possible for modules of ...
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### Example of a finitely generated module with submodules that are not finitely generated

I'm looking for an example of a finitely generated module with submodules that are not finitely generated. I've found a similar question dealing with group (i.e. an example of a finitely generated ...
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### Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?

If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a commutative ring, does it follow that $m \geq n$? This is obviously true for vector spaces over a field, but how would one show ...
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### If $M$ is an artinian module and $f$ : $M$ $\mapsto$ $M$ is an injective homomorphism, then $f$ is surjective

If $M$ is an artinian module and $f$ : $M$ $\mapsto$ $M$ is an injective homomorphism, then $f$ is surjective. I somehow found out that if we consider the module $\mathbb Z_{p^{\infty}}$ denoting ...
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### Why is the localization of a commutative Noetherian ring still Noetherian?

This is an unproven proposition I've come across in multiple places. Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian. Why is this? ...
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### 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 ...
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### Permutation module of $S_n$

Let $G=S_n$ and let $V$ be the permutation module of $G$ with basis $\{x_1,\cdots,x_n\}.$ Let $\lambda, \mu \in \mathbb{C}$ to allow one to define a $\mathbb{C}G$-homomorphism $\rho:V \to V$ by \...
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### Is $R/N(R)$ a faithfully flat $R$-module?

I'm studying recently faithfully flat modules and I'd like to know the following: Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of ...
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### Finitely generated projective module

Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free? What I know about a finitely generated projective module $M$ over a PID $R$ is ...
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### The connection between decomposition of $\ker(f_\phi g_\phi)$ and the Chinese remainder theorem

Original problem Suppose $K$ is a field, $f,g\in K[x]$ and $\gcd(f,g)=1$. $V$ is a linear space based on $K$ and $\phi\in\operatorname{End}(V)$. Notation $f_\phi$ and $g_\phi$ denote linear ...
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### Projective and injective modules; direct sums and products

I need two counterexamples. First, a direct sum of $R$-modules is projective iff each one is projective. But I need an example to show that, “an arbitrary direct product of projective modules need ...
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### Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$? [closed]

Here $A$, $B$ and $C$ are $R$-modules. Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$? And what if $B$ and $C$ are finitely generated?
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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 $\mathrm{Hom}_i(M,N)$ ...
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### Isomorphism of an endomorphism ring, how can $R\cong R^2$?

I'm aware that over a commutative ring, two free finitely generated modules of finite rank are isomorphic if and only if they have the same rank. I'm trying to understand a curious example of why ...