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

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0
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1answer
21 views

Does this stand for $\text{Hom}_R(R/I, R/I)$ ?

Let $R$ be a commutative ring with unit and $M$ be a $R$-module. We have that $\text{Hom}_R(R,M)\cong M$ as $R$-modules and $\text{Hom}_R(R,R)\cong R$ as rings. Do we have then also that $\text{...
0
votes
3answers
43 views

Is $Z/mZ\otimes Z \cong Z/mZ$?

I'm reading a Homological Algebra book that states this in some point without proving. I was trying to prove it and it seems to me that the first module is infinite and the second is not.
0
votes
1answer
44 views

Relationship between modules and maximal ideals of a commutative ring [closed]

Let $A$ be an integral domain, $M$ an $A$-module, and $m\in M$. Now for all maximal ideals $\mathfrak{m}$ there exists an $n\notin \mathfrak{m}$ such that $nm=0$. Why does this mean that $m=0$?
2
votes
1answer
26 views

If $I$ is finitely generated nilpotent and $R/I^{n-1}$ is noetherian then $R$ is noetherian

If $I$ is a finitely generated ideal of a commutative ring $R$ with $1$ such that $I^n = \{0\}$ and $R/I^{n-1}$ is noetherian, then $R$ is also noetherian. I don't know what I should do. If I can ...
0
votes
1answer
30 views

Kernel of the map given by $j(m)=1\otimes m$ is in torsion module

Let $R$ be an integral domain, $K$ its field of fractions and $M$ an $R$-module. I want to show that the kernel of the map $j:M\rightarrow K\otimes M, m\mapsto 1\otimes m$ is contained in the torsion ...
2
votes
1answer
81 views

Question regarding matrices with same image

Let $A$ and $B$ be $m\times n$ matrices over $\mathbb{Z}$ such that image($A$) = image($B$), where $A$ and $B$ are considered as maps $\mathbb{Z}^n \rightarrow \mathbb{Z}^m$. Does there exist an ...
6
votes
2answers
411 views

Some exact sequence of ideals and quotients

I saw an exact sequence of ideals $$0 \rightarrow I \cap J\rightarrow I \oplus J \rightarrow I + J \rightarrow 0$$In this sequence, maps are ring homomorphisms or module homomorphisms? And how the ...
4
votes
0answers
88 views

Testing if a submodule is free

This is hopefully a very simple question. In Gröbner Bases in Commutative Algebra by Ene and Herzog, I find the Problem 4.11, which says ($S$ here is a polynomial ring over a field $K$, $S=K[x_1\ldots ...
2
votes
2answers
459 views

Proving that the length of the sum and intersection of two finite length modules is finite.

I am trying to solve the following question: Let $M_1, M_2 \subset M$ be finite length submodules of a module $M$ (where $M$ need not be of finite length). Show that $ \ M_1 + M_2 = \{x_1 + x_2 \in M ...
0
votes
0answers
50 views

Show that it is equal to $\ker f \oplus \text{Im}f$ [duplicate]

Let $R$ be a commutative ring with unit and $M$ be a $R$-module. I want to show that if $f:M\rightarrow M$ is a $R$-module homomorphism such that $f^2=f$ then $M=\ker f \oplus \text{Im}f$. $$$$ ...
-1
votes
1answer
54 views

The endomorphism ring is a field [closed]

Let $R$ be a commutative ring with unit and $M$ be a $R$-module. I want to show that the endomorphism ring $\text{End}_R(M)=\text{Hom}_R(M,M)$ of a simple $R$-module is a field. $$$$ We have ...
1
vote
1answer
37 views

Show that it is equal to $\text{Im}f\oplus \ker g$

Let $R$ be a commutative ring with unit and $M$ be a $R$-module. I want to show that if $f:M\rightarrow N$ and $g:N\rightarrow M$ are $R$-module homomorphisms such that $gf=1_M$ then $N=\text{Im}f\...
0
votes
0answers
22 views

Direct summand of module

Let $S$ be a commutative (associative) ring with $1$, which is an extension of a commutative ring $R$ (with same $1$). Suppose that $S$, as an $R$-module, is finitely generated projetive and faithful. ...
0
votes
1answer
34 views

Annihilator - Product of cyclic groups

Let $M$ be the abelian group, i.e., a $\mathbb{Z}$-module, $M=\mathbb{Z}_{24}\times\mathbb{Z}_{15}\times\mathbb{Z}_{50}$. I want to find the annihilator $\text{Ann}(M)$ in $\mathbb{Z}$. $$$$ $$\...
3
votes
0answers
20 views

$R=\{ m+nr\sqrt{2} \mid m,n \in \Bbb Z \}$ and $I_{a,b}=\{ ma+n(b+r\sqrt{2}) \mid m,n \in \Bbb Z \}$

Let $r$ be a natural number and $R=\{ m+nr\sqrt{2} \mid m,n \in \Bbb Z \}$. We can show that $R$ is a subring of the ring $\Bbb Q [\sqrt{2}]$. My questions are as follows: $(1)$ Suppose that a ...
27
votes
5answers
5k views

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 ...
0
votes
1answer
50 views

Each cyclic $R$-module is isomorphic to an $R$-module of the form $R/J$ [closed]

Let $R$ be a commutative ring with unit. I want to show that each cyclic $R$-module is isomorphic to an $R$-module of the form $R/J$ where $J$ is an ideal of $R$. $$$$ Could you give me some ...
6
votes
3answers
323 views

Why is (one of) the criteria for a Noetherian module not implied by Zorn's lemma?

One of the equivalent definitions of a Noetherian $R$-module is Every nonempty collection of $R$-submodules has a maximal element under inclusion so, what I'm wondering is, why is this not an ...
0
votes
1answer
39 views

What's the importance of invariant factors?

I understand why the following theorem holds: If $R$ is a PID and $M$ is a finitely-generated torsion $R$-module, then there exist $q_1,...,q_s$, non-invertible elements of $R$, such that $q_i \...
2
votes
0answers
19 views

Compatibility of Yetter-Drinfeld modules.

Let $H$ be a Hopf algebra. A Yetter-Drinfeld module over $H$ is a triple $(V, \cdot, \delta)$, where $\cdot : H \otimes V \to V$ , $\delta : V \to H \otimes V$ are actions and coactions respectively, $...
0
votes
2answers
20 views

Let $S$ be a subring of a commutative Noetherian ring $R$. Then $R$ is finitely generated as an $S$-module? [closed]

Let $S$ be a subring of a commutative Noetherian ring $R$. Then how can I show that $R$ is finitely generated as an $S$-module?
2
votes
2answers
73 views

$R$ commutative ring with unity , does polynomials with unit leading coefficients of degre s from $0$ to $n$ generate all polynomials of deg $\le n$?

Let $R$ be a commutative ring with unity , consider the polynomial ring $R[x]$ , let $\mathcal P_n:=\{f \in R[x] : f=0$ or $\deg f \le n\}$ , so $\mathcal P_n$ is a finitely generated module over $R$ ....
3
votes
2answers
66 views

For $R$-modules M, is $M\cong R^{\oplus n}\otimes_RM\cong M^{\oplus n}$?

I wanted to explicitly give a bilinear map $R^{\oplus n}\times M\longrightarrow M^{\oplus n}$ when trying to prove that $R^{\oplus n}\otimes_RM\cong M^{\oplus n}$ and ended up with $(r_1,...,r_n,m)\...
3
votes
1answer
85 views

Isomorphism between endomorphism algebras

Assume that $R$ and $S$ are associative $\mathbb{C}$-algebras with unit $1_R$ and $1_S$, respectively. In addition, assume that $_RM$ is a simple left $R$-module and $_SN$ is a simple left $S$-module. ...
1
vote
1answer
44 views

A free module over $\Bbb C[x]$

Define the square matrix $A$ over $\Bbb C[x]$ by $$ A:= \begin{pmatrix} x & -2x \\ -x^{2} & x^{3}-x \end{pmatrix}. $$ Prove that the submodule $AV=\{ Av \mid v \in V \}$ of the $\Bbb C[...
3
votes
0answers
20 views

Existence of alternative basis element in free module over a PID

first question on stack exchange, please let me know if I have made any errors with formatting or in general! :) Let $f_1,f_2, ...,f_s$ be a basis of a free module $V$ over a PID $R$. Suppose that $f=...
1
vote
0answers
27 views

Make a vector space into a module on ring of polynomials

I'm teaching myself about modules and I came across this exercise that I'm having trouble with (found here: http://www.math.uiuc.edu/~r-ash/Algebra/Chapter4.pdf). The question asks: Let $T$ be a ...
1
vote
3answers
55 views

A conceptual question in ring theory?

What is the main(conceptual) difference between an ideal of a ring and a submodule over a ring?
0
votes
2answers
48 views

For all $R$-modules $N$, $R\otimes_R N\cong N$

Why is this so? This statement is from Aluffi's book Algebra: Chapter 0 and seemingly so trivial that it deserves no proof. So working with the universal property of tensor products, how exactly does ...
0
votes
1answer
17 views

What is the reduced norm map?

This is a basic question about the reduced norm homomorphism. Let $A$ be a central simple $K$-algebra and $P$ a f.g. projective $A$-module. I know that $\operatorname{End}_A(P)$ is also a central ...
0
votes
0answers
10 views

The reduced norm map $\operatorname{Nrd}: K_1(A)\to K^\times$

Let $K_1(A)$ be the Grothendieck $K_1$-group of the category of finitely generated projective $A$-modules where $A$ is a central simple $K$-algebra. I'd be grateful if someone could tell me if the ...
0
votes
1answer
25 views

Why $S\cong A/I$?

Let $A$ a $\mathbb K-$algebra of finite dimension where $\mathbb K$ is an algebraically closed field and let $S$ a simple $A-$module. In the proof of the Schur lemma, it says that since $S\cong A/I$ ...
1
vote
1answer
44 views

$A \oplus M$ is Noetherian iff $A$ is Noetherian and $M$ is a f.g. $A$-module

Let $A$ be a commutative ring and $M$ an $A$-module. Define a multiplication on the additive group $B:=A \oplus M$ by $$(a,x)(b,y) := (ab,bx+ay) \ \ \ (a,b \in A,\ x,y \in M).$$ An easy computation ...
2
votes
1answer
35 views

Minimum cardinality module for a fixed finite ring

Let $F$ be a finite field and $k$ be a positive integer. Let $M_k(F)$ denote the ring of $k\times k$ matrices. $M_k(F)$ is an $M_k(F)$-module with matrix multiplication, and $F^k$ is an $M_k(F)$-...
1
vote
3answers
95 views

$\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}$ is isomorphic to the complex numbers

I am new to tensor poducts (of modules over a commutative ring with identity) and need to understand the following example to continue with the actual exercises in my material. Namely, I need to ...
1
vote
0answers
28 views

Does $\phi: A \otimes \mathbb{Q} \to B \otimes \mathbb{Q}$ surj. imply that for $b \in B$, $b = n \phi(a)$ for some $n \in \mathbb{Z}$, $a \in A$?

Let $A$ and $B$ be abelian groups. Suppose that we have a morphism $\phi: A \to B$ and that $\phi \otimes_\mathbb{Z} \mathbb{Q}: A \otimes_\mathbb{Z} \mathbb{Q} \to B \otimes_\mathbb{Z} \mathbb{Q}$ is ...
0
votes
1answer
56 views

Projective module with non-zero annihilator [closed]

Let $M$ be a projective module. Suppose $\operatorname{Ann}_{R} \left(M \right) \neq 0$, where $\operatorname{Ann}_{R} \left( M \right) =\{r\in R : mr = 0, \ \forall m \in M \}$. Then there exists an ...
1
vote
3answers
35 views

Annihilator of modules [duplicate]

If $A$ is an $R$-module, I am having difficulty proving that $A$ is also a well-defined $R/ann(A)$-module with $(r+ann(A))a=ra$.
1
vote
1answer
27 views

$M_{1} \oplus M_{2}$ is a cyclic $A$-module $\iff \rm{Ann}(M_1)+\rm{Ann}(M_2)=A$ [duplicate]

Let $A$ be a commutative ring with an identity element $1$. An element $x$ in an $A$-module $M$ is called cyclic if $Ax=M$. An $A$-module which has a cyclic element is called cyclic $A$-module. Let $...
0
votes
0answers
15 views

Endowing module with algebra

Given a module $M$ over a ring $R$, is it possible to endow $M$ with an operation $M^{2} \to M$ that turns $M$ to an algebra that is not the trivial $m n \equiv \mathbf{0}$ identically? So far I have ...
0
votes
1answer
58 views

Same kernels for homomorphisms of free modules

Let $f: R^n \rightarrow R^m$ be an isomorphism of free $R$-modules ($R$ commutative with unity) and $\pi_1: R^n \rightarrow R^n/\mathfrak m^n$, $\pi_2: R^m \rightarrow R^m/\mathfrak m^m$ the canonical ...
1
vote
0answers
32 views

Let a,b have the same divisor (content) in an integral domain A. When can I deduce $a/b\in A^\times$?

Given a Noetherian integral domain A and a finitely generated torsion A-module M, we can define the divisor, or content, of M to be $div(M)= \sum_{P, ht(P)=1} \ell(M_P) [P]$, where the sum ranges ...
-2
votes
1answer
24 views

Module is isomorphic to a direct sum of modules, then the length(M)=sum length(M_i) [closed]

Is there any proof (or even a counterexample) for: If $M\tilde{=} M_1\oplus ... \oplus M_n$, then it follows for the finite length $l(M) = l(M_1)+...+l(M_n)$. (Modules of a commutative ring with ...
1
vote
1answer
127 views

Subring of a commutative Noetherian ring

We know that it's possible subring of a commutative Noetherian ring is not Noetherian (for example: Subring of a finitely generated Noetherian ring need not be Noetherian?). But if $S$ be a ...
2
votes
2answers
71 views

Corollary to Lemma of Nakayama

In Matsumura's Commutative Algebra there is the following Corollary to the Lemma of Nakayama: Let $A$ be a ring, $M$ an $A$-module, $N$ and $N'$ submodules of $M$, and $I$ an ideal of $A$. Suppose ...
2
votes
1answer
64 views

A better way of showing the set of all $m\times n$ matrix is an $R$ module.

Let $R$ be a ring and $m$ and $n$ be any positive integers. For any $a\in R$ and $A=(a_{ij})\in M_{m\times n}(R)$ define $aA=(aa_{ij})$. Then prove that $M_{m\times n}(R)$ is an $R$-module. Edited: ...
1
vote
1answer
39 views

Property of free modules.

Let $R$ be a PID. I want to show that if $P$ is a finitely generated left $R$-module and $P$ is isomorphic to $R^n$ and $R^m$ (as $R$-modules) for $n,m$ in the natural numbers, then $n=m$. I was ...
3
votes
1answer
113 views

How can I prove commutative rings are stably finite?

I have a proof that shows that a stably finite ring (i.e. one where $AB=I\iff BA=I$ for any two square matrices $A,B$ with entries in the ring) has the Invariant Basis Number (IBN). Trouble is, it is ...
0
votes
1answer
22 views

The existence of a non-split composition series in a indecomposable module

Assume that $R$ is a ring with unit and $M$ is a indecomposable left $R$-module with finite length. That is, $M$ has a composition series. Is it true that there is a composition series $$\begin{...
1
vote
1answer
31 views

If every cyclic $R$-module is projective, then $R$ is semisimple.

If every cyclic $R$-module is projective, then $R$ is semisimple. I know how to prove this if the definition of cyclic module is $M=Rm$ where $m$ is an element of $M$. The problem is, we defined in ...