# Questions tagged [modules]

For questions on modules over rings.

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### Intersection of (sub-)modules under Laurent and formal rings

I have a (hopefully quick) question regarding an intersection of two tensor modules. Let K be a field and $A,B$ finitely-generated modules over the Laurent series $K((X))$. Let $\tilde{A}$ be a (...
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### Lifting module homomorphisms imposing conditions on characteristic polynomials

Suppose that we are in the setting described in the first two paragraphs of this MSE post. My question wants to deal with an instance of the study of the amount of freedom that the choice of the ...
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### Primitive group rings and endomorphism rings

It is known that, for any group $G$, there exists a group $H$ containing $G$ such that the group ring $F[H]$ for some field $F$ is primitive, see Formanek, Edward; Snider, Robert L., Primitive group ... 36 views

### Extending $G$-closed sets to permutation bases of a permutation $RG$-module

I'm curious if there are any papers or results about the following scenario: Let $R$ be a commutative ring (I'm interested in particular in the $R = \mathbb{Z}$ case, but fields are okay too), $G$ a ...
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### Extending primary decomposition to submodule

Let's assume (for simplicity) that $R$ is a Noetherian ring and $M$ is a finitely generated $R$ module. Suppose we have submodules $N' \subset N \subset M$, and that we are given a primary ...
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### Surjectivity of natural map of rings

$\DeclareMathOperator\Hom{Hom}$Let $A$ be an integral domain and $P$ be a prime ideal in $A$. We denote $B=A/P$ then is the following natural map $$\Hom_A(P,A)\otimes_A B\to \Hom_A(P,B)$$ surjective? ...
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### $A^2$ is isomorphic to $A^{(\omega)}$, but not $A$

Is there an abelian group $A$ with $A\not\cong A\oplus A\cong A\oplus A\oplus A\oplus\cdots$ (a direct sum of countably many copies of $A$)? Edited to add: As no answers are forthcoming, does anyone ...
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### Structure of reflexive modules over regular local rings

Let $R$ be a regular local ring and $M$ a finitely generated reflexive $R$-module. When $R$ has dimension 2, then $M$ is a free $R$-module. This is discussed in Reflexive modules over a 2-dimensional ...
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### Subrings, submodules, and flatness

Let $R$ be a ring, and $S$ a subring of $R$. Let $M$ be a right $R$-module, and $N$ a right $S$-submodule of $M$. If $N$ is flat (or faithfully flat) as a right $S$-module, does it then follow that ...
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### Finite-exponent abelian groups

Let $G$ be an abelian group and $G=\bigoplus_{i=1}^t{{\Bbb{Z}}_{p_i}^{n_i}}^{(\Lambda_i)}$ where each $\Lambda_i$ is a set (at least one of $\Lambda_i$ is infinite). Since $G_{\Bbb{Z}}$ is a finite-...
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### Relationship between vector bundles and modules

THE GROTHENDIECK RING IN GEOMETRY AND TOPOLOGY - M.F. ATIYAH §1. The Grothendieck ring in homotopy theory I am going to be talking about vector bundles, i.e. fibre bundles with fibre a vector space ...
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### Does the choice of the algebraically closed field of characteristic $p$ have influence on the module category?

Let $G$ be a finite group and $p$ be a prime number dividing $|G|$. Let $k$ be the algebraic closure of $\mathbb{F}_p$. Let $K$ be another algebraically closed field of characteristic $p$ which is not ...
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### A Galois connection arising from discussion concerning flat module and pure exact sequence

There is some sort of symmetry in the definition of flat module and pure short exact sequence which can be made precise as follows. Let $R$ be a ring (with unit), $\mathcal{R}$ be the class of all ...
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### $\operatorname{Ext}$-group in the category of modules versus in the subcategory of finitely generated ones

I am trying to refine my understanding of derived categories. Let $\text{Mod}_R$ and $\text{Mod}^f_R$ be respectively the categories of modules and finitely generated modules over a Notherian ring $R$ ...
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### A bi-variate polynomial interpolation question

Let $R$ be a commutative unital ring, and $R^{m\times k}$ denote the set of $m\times k$ matrices with entries from $R$. A matrix $U\in R^{m\times m}$ is elementary if $U$ is obtained from the identity ...
1 vote
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### Prove that $f(M)=f^2(M)$ implies $f(M)$ is a direct summand of $M$ whenever $\text{End}_R(M)$ is a reduced ring

Let $M$ be a right $R$-module with the property that every homomorphism $\gamma:Sf\to M, f\in S=\text{End}_R(M)$, extends to $S\to M$. If $S$ has the property $f^2=0$ implies $f=0$ for every $f\in S$...
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### Absolutely irreducible representation and splitting field

Let $A$ be a finite-dimensional algebra over a field $F$. A representation $M$ of $A$ is called absolutely irreducible if $M\otimes_FE$ is irreducible as a representation of $A\otimes_FE$ for all ...
Let $k$ be a field, $A$ a finite dimensional $k$-algebra, $X$ a finite dimensional indecomposable (left) $A$-module and $M$ an infinite dimensional (left) $A$-module. Suppose further we have an ...
Let $k$ be a field and $A$ a finite dimensional $k$-algebra. Given a sequence of inclusions $M_1 \subseteq M_2 \subseteq \dots$ of $A$-modules consider the direct limit $M:= \bigcup_{i=1}^\infty M_i$. ...