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

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0answers
17 views

Is this map $\mathbf{Z}$-bilinear?

Let $M=\mathbf{Z}[C_n]$ have a $\mathbf{Z}$-basis $\{e^1,\ldots,e^{2n-1}\}$, then I want to show the following map is $\mathbf{Z}$-bilinear. $\varphi:M\times M\to\mathbf{Z}$ ...
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3answers
46 views

Localization module is nonzero

Let $R$ be a commutative ring, when $\mathfrak p$ is a prime ideal, there is the localization $M_{\mathfrak p}:=S^{-1}M$, where $S=R\setminus\mathfrak p$. Show: If $M$ is a nonzero $R$-module, ...
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1answer
50 views

Does a homomorphism $x^k\mapsto k$ exist?

If $M=\mathbf{Z}[G]$ is the group considered as a module over itself, and with $\mathbf{Z}$-basis $\{1,x,\ldots,x^{n-1}\}$, is it possible to construct a module homomorphism $\varphi:M\to\mathbf{Z}$ ...
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0answers
14 views

When is the rank of a general tensor 1?

Suppose we have two modules $M$,$N$ with $\mathbf{Z}$-bases, and take the tensor product $M\otimes_\mathbf{Z} N$. If I use Bourbaki's definition that the rank of a general tensor T is defined to be ...
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2answers
36 views

If $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules and if $M$ is a maximal ideal of $R$ then how can I show that image of $M{^m}$ is $M{^n}$?

If $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules and if $M$ is a maximal ideal of $R$ then how can I show that image of $M{^m}$ is $M{^n}$? Background: I was trying to prove that if $R{^m}$ is ...
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2answers
28 views

Semisimple modules and the radical

I don't need a proof, but can someone tell me whether it is true that for all $A$-modules $V$ we have that $V/\text{rad}V $ is semisimple, where we define $\text{rad} V$ as the intersection of all ...
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1answer
22 views

A property about quasi-primary modules

It is a fact that any discrete valuation domain $R$ has the property "P" that any proper submodule $N$ of any $R$-module $M$ is quasi-primary, in the sense that $\operatorname{rad}(N:M)$ is a prime ...
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1answer
29 views

Equivalent definitions of an algebra over a ring

I have seen two different definitions from several sources of an algebra over a ring. From Wikipedia: Let R be a commutative ring. An R-algebra is an R-Module A together with a binary operation ...
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0answers
34 views
+50

Basis of the ring $B=End_R(R^{(\mathbb N)})$

Let $B=End_R(R^{(\mathbb N)})$. Define $u,v \in B$ as $$u(e_{2_{i+1}})=0, u(e_{2_i})=e_i$$$$v(e_{2_{i+1}})=e_i,v(e_{2_i})=0$$ Prove that $\{u,v\}$ is a basis of $B$ as a $B$-module. I've already ...
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1answer
39 views

Can you always construct a map $A\otimes B\to A\times B$?

Suppose we have two $R$-mods $A,\,B$. Can we always construct a homomorphism $A\otimes B\to A\times B$?
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1answer
21 views

Question on Wedderburn components of $\mathbb C[G]$

$G$ is a finite group. Wedderburn's theorem says that $\mathbb{C}G\cong R_1 \times\cdot \cdot \cdot R_r$ as rings where $R_i=M_{n_i}(D^i)$ is the ring of $n_i\times n_i$ matrices over a division ring ...
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1answer
32 views

$(2,1+\sqrt{-5}), (1-\sqrt{-5},2)$ generate the $\mathbb Z[\sqrt{-5}]$-module $\langle 2,1+\sqrt{-5} \rangle \times \langle 2,1+\sqrt{-5} \rangle$

Ok, boring question here (I guess, at least). Let $R=\mathbb Z[\sqrt{-5}]$. Let $M=\langle 2,1+\sqrt{-5} \rangle$ the $R$-module generated by $2$ and $1+\sqrt{-5}$. I am asked to show that $M \times M ...
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1answer
13 views

Area between $y=2+|x-1|$ and $y=-\frac{1}{5}x+7$

Question 17, page 448, from Anton 8th. The question asks for the area, and the answer is 24. Now, I did draw the graph, found the points where both functions touch each other by $f(x) = g(x)$: ...
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0answers
16 views

direct sum of modules and generator subset

I am trying to solve the following problem: Let $(M_i)_{i \in I}$ be an infinite family of non zero modules and $S$ a system of generators of $\bigoplus_{i \in I}M_i$. Prove that the cardinal of $S$ ...
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1answer
20 views

Characterize semisimple rings with a unique maximal ideal

Problem Characterize the semisimple rings $R$ that contain a unique maximal ideal. I am not so sure what to do here. I know that a ring $R$ is semisimple if and only if all $R$-modules are ...
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1answer
26 views

Prove or disprove statements about modules

I am trying to determine if the following statements are true or false (i) There are free modules with non zero elements $x$ such that $\{x\}$ is linearly dependent. (ii) There are non free modules ...
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0answers
19 views

Is the projection of two isomorphic direct sum of modules isomorphic? [closed]

If A and B are R-modules, and if A x A is isomorphic to B x B, does this imply that A is isomorphic to B?
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0answers
23 views

Permutation modules and their vector space dimensions

I'm given a field $k$, a finite group $G$ and a set $S$ which $G$ acts on transitively. I'm then told to consider the permutation module $M = kS$. My first problem is understanding what the ...
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1answer
26 views

Krull-Schmidt theorem and internally cancellable modules?

According to this lecture notes (in Lemma2.1) the statement $Ae\simeq Ae^{\prime}\to A(1-e)\simeq A(1-e^{\prime})$ is true for finite dimensional algebras by using Krull-Schmidt theorem. Can anyone ...
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1answer
12 views

Statements about modules (generators and linearly independent sets)

I am trying to prove or disprove with a counterexample the following statements: (i) From every set of generators of a module $M$ one can extract a basis. (ii) Every linearly independent subset of a ...
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0answers
22 views

Presentation of a module by generators and relations

Let $R:=\mathbb C[T]$. Match the $R$-module with the presentation by generators and relations. $\bullet$$R$-modules: $M:=\mathbb C[T,T^{-1}]$ (Laurent Polynomials)$\qquad$$N:=\mathbb ...
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2answers
70 views

$k[x]/(x^n)$ module with finite free resolution is free

How to show a $k[x]/(x^n)$ module with finite free resolution is free? Suppose we have a exact sequence $k[x]/(x^n)^{\oplus n_1}\to k[x]/(x^n)^{\oplus n_{0}}\to M\to 0$, how do we get ...
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0answers
25 views

Are matroids really a generalization of independence in vector spaces?

The axioms of matroids are (Wikipedia): A finite matroid $M$ is a pair $(E,\mathcal{I})$, where $E$ is a finite set (called the ground set) and $\mathcal{I}$ is a family of subsets of $E$ (called ...
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0answers
10 views

Hopfian and Co-Hopfian Modules.

Can someone tell me why we want to study this class of modules? Let $M$ be a $R-$module. Wa say that : 1- $M$ is a Hopfian module, if every epimorphism of $M$ is a monomorphism. 2- $M$ is a ...
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2answers
30 views

An ideal which is not finitely generated

Let $K$ be a field and $A=K[x_1,x_2,x_3,...]$. Prove that the ideal $I:=\langle x_i: i \in \mathbb N\rangle$ is not finitely generated as $A$-module. I have no idea what can I do here, I mean, ...
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0answers
21 views

Semisimple algebras

In general, the tensor product of two algebras is not semisimple. Let $k$ be a ground field and let $A$ be a direct product of finite dimensional division $k$-algebras. Let $R=k^{n}$ where $n$ is ...
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0answers
27 views

Locally cyclic module exercise

An $A$-module $M$ is locally cyclic if every submodule of $M$ of finite type (finitely generated) is cyclic. (i) Show that every submodule of a locally cyclic module is locally cyclic. (ii) Prove ...
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1answer
11 views

Module of finite type has a minimal system of generators

I am trying to show the following statement: Every module of finite type, i.e. finitely generated, has a minimal system of generators (a minimal system of generators $\mathcal S$ is a system of ...
2
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1answer
109 views

Yoneda implies $\text{Hom}(X,Z)\cong \text{Hom(}Y,Z)\Rightarrow X\cong Y$??

I came across a result on the page Theorems implied by Yoneda's lemma? which said that Yoneda's Lemma implies the isomorphism of the title; namely, if we have $\text{Hom}(X,Z)\cong ...
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1answer
20 views

Under what conditions can I expect the restriction of scalars functor to preserve tensor products

Suppose I have the canonical injection $i:H\hookrightarrow G$. Evidently I can induce the map on modules which restricts scalars from $\mathbf{Z}[G]$ to $\mathbf{Z}[H]$; that is, ...
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1answer
61 views

A counterexample to $Ae\simeq Ae^{\prime}\to A(1-e)\simeq A(1-e^{\prime})$

Let $e,e^{\prime}$ be two idempotents in $k$- algebra $A$ ($k$ is a field) . Then my guess $Ae\simeq Ae^{\prime}$ (as a left $A$- module) does not imply $A(1-e)\simeq A(1-e^{\prime})$ in general, ...
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0answers
32 views

What is the tensor product of Z/10Z with Z/12Z? [duplicate]

I've just met tensor products of modules in part of my self-study and as a concrete exercise I've been asked to calculate $\mathbb{Z}/(10){\otimes}_{\mathbb{Z}} \mathbb{Z}/(12)$. Short of writing out ...
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2answers
21 views

Questions related to the concept of $k$-algebras

I am reading about modules and some days ago I've worked on some exercises related to $k$-algebras. The definition I've seen of $k$-algebra is that it is a field $k$ and a ring $A$ together with a ...
3
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1answer
36 views

Short exact sequence of modules

I am trying to show that if we have the following left splitting short exact sequence of $R-$modules: $0 \rightarrow M \stackrel{f} \rightarrow N \stackrel{g} \rightarrow S \rightarrow 0$ then there ...
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2answers
35 views

Why $\mathbb Z/ 2 \mathbb Z$ is not a free module?

I am reading some abstract algebra notes about free modules. It says that not all modules are free and the example to illustrate this is $\mathbb Z/ 2\mathbb Z$ (as a $\mathbb Z$-module) is not a free ...
5
votes
2answers
54 views

The rationals as a direct summand of the reals

The rationals $\mathbb{Q}$ are an abelian group under addition and thus can be viewed as a $\mathbb{Z}$-module. In particular they are an injective $\mathbb{Z}$-module. The wiki page on injective ...
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1answer
14 views

Morphism of $k$-algebras between abelian group of $n \times n$ matrices and $m \times m$ matrices

Problem Let $k$ be a field and $f:M_n(k) \to M_m(k)$ be a morphism of $k$-algebras ($n,m \in \mathbb N$). Prove that $n$ divides $m$. I have no idea what to do here, I thought that maybe I should ...
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1answer
31 views

Definition of Tensor Product of Modules

I am really struggling to understand several parts of the definition of tensor product given in my lecture notes: Definition of the tensor product *Denote by L the free A-module with a basis ...
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2answers
44 views

Is $R\otimes_R M\cong M$ when $R$ not necessarily commutative?

Suppose $R$ is not commutative. I am guessing in general that $R\otimes_R M\cong M$ fails to be true. However, are there any cases where this is still true? For instance group rings for a ...
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3answers
27 views

Definition of direct sum of modules?

I have just started studying modules, and am trying to figure out the definition of the direct sum of modules but I'm having trouble since different sources seem to give different definitions, for ...
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2answers
128 views

Vector space as simple $K[x]$-module

I am trying to solve the problem: Let $V$ be a vector space and $T$ a linear transformation $T:V \to V$. Let $(V,T)$ be a $K[x]$-module. Show that $(V,T)$ is simple if and only if $V$ is finite ...
2
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2answers
83 views

Why does $M\otimes k(\mathfrak{m})=M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m}$? (From Matsumura, proof of Theorem 4.8.)

Matsumura's Commutative Ring Theory, proof of Theorem 4.8, page 27, says: Let $A$ be a ring, $M$ a finite $A$-module, and $\mathfrak{m}$ a maximal ideal. If ...
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2answers
24 views

Linear independence and grammar

Let $A$ be a commutative ring. Normally, the linear independence is a property of a subset or a family elements of an $A$-module. But one often sees statements like "$v$ and $w$ are linearly ...
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2answers
41 views

$M$ is a simple module if and only if $M \cong R/I$ for some $I$ maximal ideal in $R$.

I am trying to show the following statement (taken from Rotman's Advanced Modern Algebra): Let $M$ be an $R$-module. Then $M$ is a simple module if and only if $M \cong R/I$ for some $I$ maximal ...
4
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1answer
73 views

Rank of projective module defined as the smallest $n$ such that $P$ is a direct summand of $R^n$

Over a commutative ring $R$, the rank of a projective module $P$ is defined by looking at the map $\text{rank}(P) : \text{Spec}(R) \rightarrow \mathbb{N}_0$ given by $\mathfrak{p}\mapsto ...
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0answers
19 views

A dual to essentially finitely generated notion

It is known that a (right) module $M$ over a ring $R$ has finite uniform dimension if and only if it is "essentially finitely generated", in the sense that there exists a finitely generated submodule ...
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0answers
60 views

Using the universal property of tensor product to show that $(\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}$ [duplicate]

In the algebra lecture i need to solve the following exercise Use the universal property of the tensor product to show that $$(\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong ...
0
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1answer
30 views

Problem related to modules and k-algebras

I am trying to do the following exercise: Prove that if $A$ is a $k-$algebra and $M$ is a module then the product $\lambda * m=\tau(\lambda)m$ ($\tau$ is the morphism from $K$ to $Z(A)$) defines on ...
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votes
1answer
44 views

Lifting a direct summand of a free module

Suppose $R$ is a commutative ring, $I\subseteq R$ a principal ideal, and we're given split short exact sequences $ R \to R^n \to R^{n-1}$ and $ R/I \to (R/I)^n \to (R/I)^{n-1}$ the first inducing ...
0
votes
1answer
25 views

A simple module

Suppose $R$ is a ring and $M$ is an $R$-module. Prove: $M$ is simple if and only if there is a left maximal ideal $m$ such that $M\cong R/m$.