Questions tagged [modules]

For questions on modules over rings.

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Group algebra of a subgroup

I think what I want to do is not very hard, but I am just not seeing it. Let $K$ be a subgroup of a finite group $H$, and let $A$ be a commutative ring with $1$. Consider the group algebras $A[K]\...
EJB's user avatar
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Exact sequences with two FL-modules

Let $R$ be a ring. An $R$-module $M$ is called FL (FP) if it has a finite resolution consisiting of finitely generated free (projective) modules. Given an exact sequence of $R$-modules, $0\to M_1\to ...
Andrei Jaikin's user avatar
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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 (...
mathieu_matheux's user avatar
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Automorphisms of matrix algebras and Picard group

This is a repost of https://math.stackexchange.com/questions/4692364/automorphisms-of-matrix-algebras-and-picard-group (asked on MSE). Notation. In what follows, $R$ is a commutative ring with $1$, $n\...
GreginGre's user avatar
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Does $M$ satisfy the descending chain conditions on $\mathbb{Z}G$-retracts?

‎Let $H$ be a subgroup of $G$‎. ‎Then a homomorphism $r:G\to H$ is said to be a retraction if the inclusion homomorphism $i:H\hookrightarrow G$ is a right inverse of $r$‎, ‎i.e‎. ‎$r(x)=x$ for all ...
M.Ramana's user avatar
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Commuting matrices and cyclic modules

Let $A, B\in M_n(\mathbb{C})$ be matrices that commute. We suppose that there exists a vector $v\in\mathbb{C}^{n}$ such that $(\mathbb{C}[A,B]).v$ generates $\mathbb{C}^{n}$. We call such a pair a ...
prochet's user avatar
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Intersection of two modules (and sub-modules) under tensors

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 (...
M-M's user avatar
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4 votes
1 answer
73 views

Property of simplicity and semi-simplicity under base change of base field

Suppose $K$ is a field of characteristic $0$ and $A$ is a $K$-algebra. Let $F$ be a field extension of $K$ and let $M$ be an $A$-module. What can we say about simplicity or semi-simplicity of $A_F$-...
amir hossein Ekhlasi's user avatar
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Proving finite presentation [closed]

Let $R$ be an integral domain, $S$ be a finitely presented $R$ algebra. Then for a flat $R$ module $M$ which is also a finitely generated $S$ module I need to show that $M \otimes_{R}T$ is a fintely ...
user443060's user avatar
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Clebsch–Gordan decomposition formula for algebraic groups

$\DeclareMathOperator\SL{SL}$There is a well-known Clebsch–Gordan decomposition formula for irreducible representations of $\SL_2$. If $V_n$ denotes the unique $n+1$-dimensional irreducible ...
dm82424's user avatar
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modules with the same associated variety

$S$ is a polynomial ring, $M$ is a finitely generated graded $S$-module, the associated variety of $M$ is $$ \mathcal{V}\left( M\right)=V\left( Ann_S \left( M \right) \right), $$ What is the ...
zhjzwlys's user avatar
4 votes
1 answer
176 views

Classifying indecomposable modules over $\mathbb{Z}/p^{2}\mathbb{Z}[\mathbb{Z}/p\mathbb{Z}]$

I'm now interested in classifying the indecomposable modules over $\mathbb{Z}/p^{2}\mathbb{Z}[\mathbb{Z}/p\mathbb{Z}]$ : the group ring of $\mathbb{Z}/p\mathbb{Z}$ over the ring $\mathbb{Z}/p^{2}\...
gualterio's user avatar
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2 answers
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Torsion of modules

Given a left module $M$ over a domain $R$, I am interested in irreducible elements $r\in R$ such that $r\cdot m=0$ for some $m\in M-\{0\}.$ I think "torsors" would be perfect name for such $...
Adam's user avatar
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8 votes
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Is the category of chain complexes a reflexive or coreflexive subcategory of the category of functors?

Let $A$ be an abelian category (you can assume additional conditions for its goodness). Let $\mathrm{Seq}(A) = \mathrm{Func}(\mathbb{Z}, A)$, where $\mathbb{Z}$ is the standard order category on ...
Arshak Aivazian's user avatar
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When do two path algebras share an underlying graph?

Suppose $Q$ and $Q'$ are two quivers. I am curious as to what relation $\mathbb{C}Q$ bears to $\mathbb{C}Q'$ when $Q$ and $Q'$ share the same underlying graph and only differ by direction. Since ...
tox123's user avatar
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For a finite dimensional $k$-Algebra $A$ does infinite representation type imply $(\text{rad}_A^{\omega})^2 \neq 0$?

Let $k$ be a field, $A$ a finite dimensional $k$-Algebra, $\text{mod}\,A$ the category of finite dimensional left $A$-modules and $\text{rad}_A$ the collection of radical morphisms in $\text{mod}\,A$. ...
kevkev1695's user avatar
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Example of a secondary representation of a module that is not a direct sum

Let $A$ be a commutative ring. An $A$-module $M$ is said to be secondary if $M\neq 0$ and for each $a\in A $, the endomorphism $\phi_a:M\to M$ defined by $\phi_a(m)=am$ for $m\in M$ is either ...
L. Xie's user avatar
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Which definitions of "local module" have gotten traction?

It seems like "local module" has been defined a lot of ways: if 𝑀 has a largest proper submodule. (This math.se post) if it is hollow and has a unique maximal submodule (Singh, Surjeet, ...
rschwieb's user avatar
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Decidability of theory of modules over a ring of finite representation type

I have read from Mike Prest's model theory for modules (London lecture note series) chapter 17 that a Ring of finite representation type has a decidable theory of modules. Here decidability was ...
Yoneda Lemma's user avatar
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Where could a paper on a unification of matrix decompositions be published?

I've got a paper which shows that when the spectral theorem (as a statement that every self-adjoint matrix can be unitarily diagonalised) is naively generalised to $*$-algebras other than the complex ...
wlad's user avatar
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Finding a particular kind of basis of subgroup of a lattice generated by non-negative part

For $\mathbf v=(v_1,\ldots,v_n)\in \mathbb Z^n$, let $\operatorname{supp}(\mathbf v):=\{j: v_j \ne 0\}$. For a subset $X$ of $\mathbb Z^n$, define $\operatorname{supp}(X):=\bigcup_{\mathbf v \in X} \...
uno's user avatar
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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 ...
Hvjurthuk's user avatar
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1 answer
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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 ...
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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 ...
Sam K's user avatar
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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 ...
Sasha's user avatar
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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? ...
tota's user avatar
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24 votes
1 answer
563 views

$A^2$ 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 ...
Pace Nielsen's user avatar
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2 votes
1 answer
130 views

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 ...
Ahmed Matar's user avatar
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1 answer
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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 ...
Jake Wetlock's user avatar
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116 views

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-...
Najmeh Dehghani's user avatar
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0 answers
96 views

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 ...
Abel 's user avatar
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7 votes
1 answer
220 views

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 ...
LSt's user avatar
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5 votes
0 answers
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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 ...
Zhenhui Ding's user avatar
2 votes
0 answers
141 views

$\mathbb Z_n[x_1^\pm,\dots,x_D^\pm]$-modules extended from $\mathbb Z_n$

Let $n$ be a positive integer and let $\mathbb Z_n=\mathbb Z/n \mathbb Z$. Consider the ring of Laurent polynomials $R=\mathbb Z_n[x_1^\pm,\dots,x_D^\pm]$. $R$-modules of the form $M=M_0 \otimes_{\...
Blazej's user avatar
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1 vote
1 answer
223 views

Product-coproduct duality

Let $T$ be a set, $R$ be a ring with $1$ and $B, S_t$ be $R$-modules $\forall t \in T$ My task is to state and prove the dual to the following statement: Given momomorphisms $j_t: S_t \rightarrow B$. ...
Igor Kharin's user avatar
5 votes
1 answer
192 views

What are the indecomposable modules over $\mathbb{F}_2(C_2\times C_2)$?

Let $C_2$ be the cyclic group of order $2$ and $\mathbb{F}_2$ the field with $2$ elements. Consider the group algebra $A:= \mathbb{F}_2 (C_2\times C_2)$. It is well-known that $A$ has infinite ...
kevkev1695's user avatar
3 votes
1 answer
161 views

Modules with special properties

$\DeclareMathOperator\End{End}$Let $A$ be a finite dimensional algebra and $M$ an indecomposable (right) module with the property that every nilpotent element of $\End_A(M)$ annihilates the socle $\...
Mare's user avatar
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0 votes
1 answer
138 views

Semi simplicity over commutative algebras over non-algebraically closed fields

I have already posted this on stackexchange I have a question: If k is an arbitrary field then is it true that if $M$ a finite dimensional $k[x, y]$ is semisimple as a $k[x]$ module and also as a $...
Subham Jaiswal's user avatar
3 votes
0 answers
100 views

Are there (co-)homological obstructions to the extendability of a homomorphism?

Let $k$ be a commutative ring and $A \subset B$ an extension of $k$-algebras. Can we associate to this extension a (co-)chain complex so that its (co-)homology $H$ allows for statement such as "...
Nicolas Cage's user avatar
2 votes
0 answers
435 views

Minimal number of generators of kernel of a matrix of polynomials over $\mathbb Q$ and $\mathbb Z$

$\DeclareMathOperator\im{im}$I have asked some related questions before on math.SE here and on MathOverflow here (answered here). This post is self-contained. Let $R' = \mathbb Q[x_1,\dotsc,x_n]$, and ...
Pranay Gorantla's user avatar
1 vote
1 answer
275 views

Constructing a free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-module using a related free resolution of a $\mathbb Q[x_1,\dotsc,x_n]$-module

I have asked a related question on math.SE here, but the notation is a bit different. As the title says, I am interested in constructing a finite free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-...
Pranay Gorantla's user avatar
1 vote
1 answer
62 views

Hilbert - Samuel multiplicity of $B$ when there is a surjection $A \rightarrow B$

The setting is: Let $A, B$ be commutative, Noetherian, local rings, $\phi:A \rightarrow B$ a surjective homomorphism. Both rings also come with surjections $\lambda_A, \lambda_B$ to a DVR $\mathcal{O}$...
JBuck's user avatar
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1 vote
1 answer
121 views

Pontrjagin dual of modules [closed]

I am not sure whether this question is appropriate to appear here. If not, I apologize for that. Given an $R$-module $M$, we define its Pontrjagin dual as $M^{\ast}=Hom_{\mathbb{Z}}(M, \mathbb{Q/Z})$. ...
ZYX's user avatar
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3 votes
1 answer
202 views

Nondegenerate pairings versus perfect pairings for finitely generated projective modules

Let $R$ be a (not necessarily commutative) ring, $M$ a left $R$-module, and $N$ a right $R$-module. We say that a pairing $$ \langle -,-\rangle:M \otimes_R N \to R $$ is non-degenerate if, for all $n \...
Adam Bondal's user avatar
0 votes
0 answers
78 views

$\operatorname{Ext}$-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 $R$ ...
Stabilo's user avatar
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1 vote
0 answers
40 views

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 ...
TimeaV's user avatar
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1 vote
0 answers
60 views

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 $f\in S$...
mariam's user avatar
  • 111
5 votes
1 answer
340 views

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 ...
Hebe's user avatar
  • 729
7 votes
0 answers
255 views

Split epimorphism of modules - does the finite case imply the infinite case?

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 ...
kevkev1695's user avatar
1 vote
1 answer
82 views

Epimorphism going out of an inverse limit into a finite dimensional module

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 $M:= \bigcup_{i=1}^\infty M_i$. ...
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