Questions tagged [modules]

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

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Subcategory determined by composition series

Suppose $A$ is an artin algebra and take the category $\operatorname{mod}A$ of finitely generated $A$-modules. Consider the following construction. Let $M$ be an $A$-module. Since $A$ is artinian, $M$ ...
sirferrum's user avatar
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Irreducible module over upper triangular matrices

Let $R$ be a ring and $M$ be an irreducible $R$ module. I want to create an irreducible $A$-module, where $$A = \begin{pmatrix}R & R\\0 & R\end{pmatrix}.$$ I defined $\bar{M}$ as $$\bar{M} = \{...
Guilherme Costa's user avatar
1 vote
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Facts about Weyl algebra

I am trying to prove a few things about the first Weyl algebra, $W = k[x,y]/(xy-yx-1)$ over an algebraically closed field $k$ with $char(k)=p>0$. In particular, I am interested in nilpotent ...
wwinters57's user avatar
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Epimorphism and Surjective Homomorphsim [duplicate]

Here is a question I came across recently: If a morphsim in Grp (category of groups) is an epimorphism, then it is a surjective group homomorphsim. I believe it boils down to show that for any group $...
JNF's user avatar
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Finitely generated ideals of integer valued polynomials

Let $\operatorname{Int}(\mathbb{Z}):= \{f(x) \in \mathbb{Q}[x] \mid f(\mathbb{Z}) \subseteq \mathbb{Z}\}$. Any element of the integer valued polynomials can be written as a $\mathbb{Z}$ linear ...
Divya's user avatar
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Chain map sequence $0 \to H(C) \to C/B(C) \xrightarrow{d} Z(C)[-1] \to H(C)[-1] \to 0$ is exact for a cochain complex of $R$-modules?

Specifically, Weibel Page 10 Exercise 1.2.7b. I think I proved Exercise 1.2.7a or that there exists an SES: $0 \to Z(C) \to C \xrightarrow{d} B(C)[-1] \to 0$ given a complex of $R$-modules $C^{\cdot}$ ...
MathCrackExchange's user avatar
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Example about endomorphism ring of an indecomposable object, which have non-trivial idempotent element in an additive category.

Recently I read some about Krull-Remak-Schmidt category. If $A$ is an additive category in which every idempotent splits, every object is the biproduct of finitely many indecomposable objects and the ...
Well's user avatar
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notation for creating formal power series and $R-module$

The following is taken from "Module Theory an approach to linear algebra" by T.S Blyth $\color{Green}{Background:}$ $\textbf{(1) Example}$ Let $R$ be a unitary ring and let $R^N$ denote the ...
Seth's user avatar
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Example 2.2 in Rotman's Homological Algebra Text

1. Example 2.2 (excerpt) 2. Questions If $x,y \in G$ are distinct, then in $kG$ we have $xy = \delta_x \delta_y = \delta_{x,y} = 0$. I think this is somehow wrong. The example says that $1 \in G$ is ...
IsaacR24's user avatar
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$M$ being the direct sum of submodules $(M_i)_{i \in I}$ is equivalent to a certain map (between $\operatorname{Hom}$-sets) being an isomorphism.

I am currently trying to prove a remark in Bosch: Algebraic Geometry and Commutative Algebra (chapter 1.4): Consider a family $(M_i)_{i \in I}$ of submodules in $M$. Then the inclusion maps $\iota_i \...
puck29's user avatar
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Trouble understanding torsion

I'm reading this paper. In definition 6.1.2 it is mentioned that $H_n(G;A) \cong \text{Tor}_n^{\mathbb{Z}G}(\mathbb{Z}, A)$ (for all $n=0, 1, 2, ...$). What exactly is meant by the right-hand side? I'...
JMM's user avatar
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Counterexample to $\hat{\mathfrak{a}}\hat{M}=\widehat{\mathfrak{a}M}$ when the base ring is not Noetherian or the module $M$ is not finitely-generated

$\def\fra{\mathfrak{a}}$Here it is proven that for $A$ a Noetherian ring, $\fra\subset A$ an ideal and $M$ a finitely-generated $A$-module and if we take $\fra$-adic completions, then $\hat{\mathfrak{...
Elías Guisado Villalgordo's user avatar
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Is this solution to show that the Ideal given by the kernel of$ f \in R \mapsto f(0,0) \in \mathbb C[X,Y]$ is not finitely generated correct?

Let $R \subseteq \mathbb C[X,Y]$ be the subring of all polynomials $f \in \mathbb C[X,Y]$ that can be written as $f = g(X)+X ·h(X,Y)$ a. Let $I \subset R$ be the kernel of the evaluation map $R \...
some_math_guy's user avatar
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1 answer
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Jordan-Holder theorem for group algebras

I'm currently studying the Jordan-Holder theorem for modules and representations of associative algebras over fields. I was wondering if there is a way to prove the Jordan-Holder theorem for finite ...
Lorenzo Ferraiuolo's user avatar
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1 answer
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The Ideal given by the kernel of map $ f \in R \mapsto f(0,0) \in \mathbb C[X,Y]$ is not finitely generated for this particular polynomial subring

In the following problem, $R$-module means left $R$-module and $R$ is a ring. I have already proven these facts that may or not be needed: -Show that an $R$-module $ M$ is finitely generated if and ...
some_math_guy's user avatar
1 vote
1 answer
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Let $p$ be an odd prime number, and let $m \ge 0$ and $N \ge 1$ be integers. Let $\Lambda$ be a free $\mathbb{Z}/p^N\mathbb{Z}$-module of rank $2m+1$,

$$ \text{Let } p \text{ be an odd prime number, and let } m \ge 0 \text{ and } N \ge 1 \text{ be integers. Let } \Lambda \text{ be a free } \mathbb{Z}/p^N\mathbb{Z} \text{-module of rank } 2m+1, \text{...
Martin's user avatar
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1 answer
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Is being finitely generated a local property

Searching on this site and others leads to lots of dicussion about localisation at multiplicatively closed subsets of the form $\{f_i^j\}_{j=1}^\infty$ where $\{f_i\}_{i=1}^n$ generate the whole ring ...
DevVorb's user avatar
  • 733
1 vote
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Not sure how to show a module homomorphism to a finitely generated module is an isomorphism.. [duplicate]

I have the question: Let $M$, $N$ be modules over a ring $R$ with homomorphisms $f, g : N \longrightarrow M$ such that $f$ is surjective and $g$ is injective. Show that: (1) $f$ is an isomorphism if $...
mad_scientist's user avatar
2 votes
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Proving a lemma about an extending $R$ module with no $M$-singular direct summands in "Extending modules" (1994)

I'm reading this book: Dung, N. V., Van Huynh, D., Smith, P. F., & Wisbauer, R. (1994). Extending modules (Vol. 313). CRC Press. In the proof of Lemma 11.1, there is a part that I can not ...
Na Man's user avatar
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1 answer
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Alternative proof for Structure theorem for finitely generated modules over a principal ideal domain

I'm thinking of an alternative proof for Structure theorem for finitely generated modules over a principal ideal domain (also called Fundamental theorem of finitely generated modules over a PID in ...
RHspqr's user avatar
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Linearly Compact Module in $R-Mod$

Definition: A module $M$ is called linearly compact if for a family of cosets $\{x_{i}+M_{i}\}_{\triangle}$, $x_{i}\in M$, $\triangle$ is a directed set, and submodules $M_{i}\subset M$ (with $M/M_{i}$...
YSA's user avatar
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Can I use Auslander-Buchsbaum formula for polynomial rings? [closed]

Let $R=k[x,y]$ be a polynomial ring over a field $k$ and set $T=R/\langle xy,y^2‎‎\rangle $‎. ‎Let $f:R\to T$ be the natural ring epimorphism‎. ‎Is $\operatorname{pd}_RT=2$? we know that $\...
pink floyd's user avatar
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When is $\mathbb Z/n\mathbb Z$ semisimple as $\mathbb Z$-module? [duplicate]

I'm looking for some characterization of $\mathbb Z/n\mathbb Z$ being semisimple as $\mathbb Z$-module. Obviously for $n$ prime this is the case, as $\mathbb Z/n\mathbb Z$ is simple under that ...
algebrah's user avatar
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Classification of 4 dimensional real associative unital algebra

I think I have a complete list for all the commutative ones, maybe with possible repeats (I did try my best to make sure none are same up to isomorphism): $\mathbb{R}^4 \simeq \begin{bmatrix}a&0&...
Leon Kim's user avatar
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1 vote
1 answer
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submodule of completely reducible module is completely reducible

I attempted to solve problems from the textbook 'Basic Abstract Algebra' by P.B. Bhattacharya, S.K. Jain, and S.R. Nagpaul, specifically Chapter 14, Section 4, but encountered difficulties in ...
N00BMaster's user avatar
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The name for a type of map between vector spaces

Is there a name for a map $f:V \to W$ between two $\mathbb{K}$-vector spaces that is not linear map but which still staisfies $$ f(\lambda v) = \lambda f(v), ~~~~~ \textrm{ for all } \lambda \in \...
Lorenzo Del Vecchiopontopolos's user avatar
1 vote
1 answer
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For $\varphi : \bigoplus_I R \rightarrow M$ the canonical projection, when is $\varphi^{-1}(m)$ finite for each $m \in M$?

For any $R$-module $M$, we can find some set of generators $S$ and construct a surjective map $\varphi : \bigoplus_S R \rightarrow M$ by sending the basis elements of $\bigoplus_S R$ to the generators ...
tolUene's user avatar
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1 vote
0 answers
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Construct a short exact sequence that is not split

I am currently studying about exact sequence. To illustrate that not all short exact sequences are split, my teacher provided an example in the homework. For every $m \geqslant2$ construct a short ...
Henry Ford's user avatar
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Question on Proof of Splitting Lemma for Modules

The setup of the question is as follows. Let the following be a short exact sequence of modules: $$ 0 \rightarrow A \xrightarrow{i} B \xrightarrow{p} C \rightarrow 0, $$ and let $r: B\to A$ be a map ...
pyridoxal_trigeminus's user avatar
3 votes
1 answer
38 views

Noetherian, self-injective ring $R$ with non-torsionless $R$-module

Let $R$ be an (associative, unital) ring. By an $R$-module we mean a left $R$-module. We call an $R$-module torsionless if it can be embedded into a direct product of the regular $R$-module $R$. I am ...
Margaret's user avatar
  • 1,055
2 votes
1 answer
67 views

Overview for properties of modules

I was wondering if there exists a nice overview for properties of modules like being cyclic, simple, semisimple, indecomposable, free, noetherian, artinian and having finite length. I'm thinking about ...
algebrah's user avatar
  • 107
6 votes
4 answers
217 views

Every $K[G]$-module is torsionless?

Let $G$ be a finite group and $K$ a field. Consider the group ring $R:=K[G]$. Let $M$ be a (left) $R$-module. Is it true that then there exists a set $S$ and an injective $R$-module homomorphism $M\...
Margaret's user avatar
  • 1,055
0 votes
1 answer
45 views

How to understand Free Module $K[x]^r=\bigoplus_{i=r}^rK[x]e_i$

How can i understand of the free module $K[x]^r=\bigoplus_{i=r}^rK[x]e_i$ where $e_i=(0,\ldots ,1, \ldots 0) \in K[x]^r $ denotes the i–th canonical basis vector of $K[x]^r$. We call $x^\alpha e_i=(0,...
Kevin Duran's user avatar
1 vote
0 answers
41 views

Every finitely generated module is sum of local modules

Note: All modules are over a $K$-algebra $A$ with $K$ a field and the underlying ring of $A$ is unital (but not necessarily commutative). Definition: A module $V$ is local if there is a maximal ...
Gargantuar's user avatar
-4 votes
1 answer
54 views

Free module over a countable set [closed]

Suppose $R$ is a commutative ring. Is $R^{\oplus \mathbb{N}}$ isomorphic to $R^{\oplus \mathbb{N} }\oplus R^{\oplus \mathbb{N}}$ as $R$-modules? If so, how do I find an explicit isomorphism? Edit: I ...
Pecfex's user avatar
  • 25
1 vote
1 answer
47 views

Why direct sum of modules admits canonical projections?

This is more of a moral (i.e. category theoretical) question. In the category of $R$-modules for a ring $R$, the product is the direct product $M=\prod_{i\in I}M_i$ with canonical projections $\pi_i\...
Gargantuar's user avatar
2 votes
0 answers
26 views

The Grassmann connection is a connection

Let $\mathcal{A}$ be a *-algebra and $p\in M_N(\mathcal{A})$ an orthogonal projection. I need to show that $\nabla=p\circ d$ defines a connection on $\mathcal{E}=p\mathcal{A}^N$, where $d$ is acting ...
Schrödinger's cat's user avatar
2 votes
1 answer
55 views

$\mathbb{Q},\mathbb{R}$ and $\mathbb{C}$-vector space isomorphism from $\mathbb{Z}$-module isomorphism.

Let $A,B$ be $\mathbb{C}$-vector space. We can view them as a $\mathbb{Z}$-module. Suppose that there is a $\mathbb{Z}$-module isomorphism $\phi$ between $A$ and $B$. Then can we have a natural $\...
PZM's user avatar
  • 21
1 vote
1 answer
40 views

rank of a free submodule of a free module of infinite rank

I am currently studying the free modules and I am stuck in the following question. Please help me. I know that if $M$ is a free module on an infinite subset $A$ over a ring $R$ (not necessarily ...
Ratanjit 's user avatar
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39 views

Understanding the free $R$-algebras.

Let $R$ be a commutative ring, $R\left< x,y \right>$ the free $R$-algebra on indeterminates $x$ and $y$. If $z_i=xy^i$, I want to show that the $R$-subalgebra generated by the elements $z_i$ is ...
Ty Perkins's user avatar
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0 answers
28 views

Doubt concerning the definition of the tensor product of modules

Let $R$ be a ring with unit and consider a right $R$-module $M$ and a left $R$-module $N$. The tensor product $(M \otimes_R N, \otimes_R )$ of $M$ and $N$, is usually defined as the quotient of the ...
Victor's user avatar
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1 vote
0 answers
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Reference Request: Extension of Scalars, Free and Projective Modules

In the comments of this question: What does it mean when the extension of scalars is free?, it is mentioned that if $ \psi: R \rightarrow S $ is a faithfully flat ring homomorphism between commutative ...
Tom Adams's user avatar
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0 answers
31 views

Automorphism group of an R-module

Let R be a commutative ring. It is well known that the automorphism group of the module $R^n$ is isomorphic to $GL_n(R)$. Is their a way to measure how much this fails for an arbitrary module M? ...
DevVorb's user avatar
  • 733
2 votes
1 answer
46 views

Indecomposable $K[T]$-modules with $\mathrm{char}(K)=2$

Let $K$ be a field of characteristic $2$. Let $G = \mathbb Z / 2\mathbb Z$. The goal is to determine all indecomposable $KG$-modules up to isomorphism. My main problem is to show that such a module is ...
Gargantuar's user avatar
0 votes
0 answers
21 views

Submodule of $\mathbb{Z}$-module with the same rank

For a $\mathbb{Z}$-module $M$ in the field of algebraic numbers, suppose that it has a $\mathbb{Z}$-submodule $N$ which has the same rank with $M$. Shall one deduce that $M=N$? Or can anyone please ...
zyy's user avatar
  • 967
1 vote
0 answers
26 views

If two integer matrices $A$ and $B$ have the same row space, $A=UB$ for some unimodular matrix $U$?

I am trying to prove that if $A$ and $B$ are matrices over the integers of the same dimension $m\times n$ and they have the same row space (that is, the set of all linear combinations with integer ...
ABC's user avatar
  • 874
0 votes
1 answer
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Functor from a filtered category to the category of R-Mod

The following result appears in Charles Weibel's book An introduction to homological algebra as lemma 2.6.14 in page 56. Lemma Let $I$ be a filtered category and let $A:I\rightarrow R\text{-mod}$ be a ...
Ziqiang Cui's user avatar
1 vote
1 answer
29 views

Example request of a finite dimensional algebra of infinite global dimension with a module with no self-extensions

I'm looking for an example of a finite dimensional algebra of infinite global dimension with a non projective module $X$ with no self-extensions, that is $\text{Ext}_A^i(X,X) = 0$ for $i>0$. The ...
Momo1695's user avatar
0 votes
1 answer
29 views

relation between support and submodules

Support of an $R$-module $M$ is defined as $\mathrm{Supp}_R(M)=\{\mathfrak{p}\in Spec(R); M_{\mathfrak{p}}\neq 0\}$. Let $M$ and $N$ two finitely generated modules over a Noetherian ring $R$ such that ...
Saeed Yazdani's user avatar
3 votes
1 answer
71 views

Infinite Product of Module categories over rings is not equivalent to Module category over infinite product of rings

I am looking for a counterexample to the following: Let $(A_i)_{i = 1}^{\infty}$ be an infinite family of rings with unity. Let $\text{Mod}A$ denote the category of right $A$ modules for a ring $A$. ...
Subham Jaiswal's user avatar

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