Questions tagged [modules]

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

Modules are abelian groups with an added notion of multiplication by elements in a ring. They generalize abelian groups, which are modules over the integers, and vector spaces, which are modules over a field.

Rigorously, a left $R$-module is defined as an abelian group $M$ paired with a ring $R$ with a binary operation from $\cdot\;\colon R\times M\rightarrow M$ satisfying the following axioms for all $m,n\in M$ and $r,s\in R$:

  1. $r\cdot(m+n)=r\cdot m+r\cdot n$

  2. $(r+s)\cdot m=r\cdot m+s\cdot m$

  3. $(rs)\cdot m=r\cdot(s\cdot m)$

If $R$ is a unital ring, we often also require that $1\cdot m=m$.

A right module is defined similarly by rewriting the axioms with the ring elements acting on the right side.

Modules often arise in the study of commutative rings and in algebraic geometry, but may appear in any investigation of the structure of a ring as a result of the Yoneda embedding which sends a ring to the category of left modules over that ring.

9674 questions
5
votes
1 answer

Definition of $R$-modules

I have a dumb question on the definition of $R$-modules. I borrow the definitions from Wikipedia. A left $R$-module $M$ over the ring $R$ consists of an Abelian group $(M,+)$ and an operation $R\times M \rightarrow M$ such that for all $r,s\in R$…
Lord Soth
  • 7,750
  • 20
  • 37
5
votes
2 answers

For $M$ a finitely generated $R-$module with $R$ a PID, then $F$ in the decomposition of $M=F\oplus M_{\text{tor}}$ doesn't have to be unique.

The question is in the title. It's really hard to "show my effort" on this, because it's just coming up with an example. My effort is... thinking about it for a while? I thought about letting $R = \mathbb{Z}$ or $\mathbb{R}[x]$ but I'm having…
Moni145
  • 2,142
5
votes
1 answer

Simple module over a basic algebra is one-dimensional

May I refer you to page $32$ of: http://www.staff.city.ac.uk/a.g.cox/LTCC/Week3.pdf Part (b) of proposition $3.2.4$. Why the fact that $A/\operatorname{rad}(A)$ is isomorphic to $k^m$ for some $m$ implies that the dimension of any simple $A$-module…
user10
  • 5,688
5
votes
3 answers

Sum is an isomorphism, finitely generated modules

Suppose $R$ is a local ring with maximal ideal $m$ and suppose $M$ and $N$ are $R$-finitely generated modules. Let $f,g: M \rightarrow N$ be an $R$-module homomorphism. If $f$ is an isomorphism and $g(M) \subset mN$ why is $f+g$ in fact an…
user6495
  • 3,957
5
votes
1 answer

Show that $Hom_R(R^n,M) \cong M^n$ for R-modules

We want to show that $Hom_R(R^n,M) \cong M^n$ for $n\in\Bbb Z_{\ge0}$ I have already shown that $Hom_R(R,M) \cong M$ by letting $f:Hom_R(R,M)\rightarrow M$ given by $f(\phi) = \phi(1)$. I showed that $f$ is bijective and is a group homomorphism,…
5
votes
2 answers

Conceptual doubt about Modules

Let $R$ be a ring. Then a left $R$- module is an additive Abelian group M along with an operation $R\times M \to M$ such that it satisfies - $r(x+y)=rx+ry$ $(r+s)x=rx+sx$ $(rs)x=r(sx)$ $1_R x=x$ where $x,y \in M $ and $r \in R$ Now my…
blabla
  • 1,104
4
votes
1 answer

Any isomorphism implies a split exact sequence?

For an exact sequence of modules $$ 0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0 $$ there is a map $\pi: B \rightarrow A$ for which $\pi \circ f = 1_A$ if and only if $A \oplus C$ is isomorphic to $B$ in such a way that $f$…
D_S
  • 33,891
4
votes
1 answer

Elementary question about localization of ideals

I'm trying to show the following: Let $I,J$ be ideals of a commutative ring (with 1) $A$ such that $I_{P}=J_{P}$ for every prime ideal P of R. Here $I_{P}$ means the localization of $I$ in $P$. Then $I=J$. Well I was thinking in using the following…
user6495
  • 3,957
4
votes
1 answer

On the Converse of Schur's Lemma for Modules

Let $R$ be a ring (with unit) and $M$ a $R$-Module. From Schur's Lemma (for Modules) we know that, if $M$ and $N$ are simple $R$-modules, every $\phi \in Hom_R(M,N)$ is either the zero morphism or an isomorphism, thus $Hom_R(M,N)$ is a division…
Jonas Gomes
  • 3,087
4
votes
1 answer

If $A/I_i$ is noetherian for all $i$, and $\cap I_j=\{0\}$, is $A$ noetherian?

I want to show the following: Let $I_{1},...,I_{n}$ be ideals of a commutative ring (with 1) $A$ such that $\bigcap_{i=1}^{n} I_{i}$ is the zero ideal. If each quotient $A/I_{i}$ is a Notherian ring show that $A$ is a Notherian ring. My work: Let…
user6495
  • 3,957
4
votes
0 answers

Normal Form problem (in a module over a PID)

Let $A,B$ be $n\times n$ matrices with entires in a PID $D$ and $\det AB\neq 0$. Suppose diag$\{a_i\}$, diag$\{b_i\}$, and diag$\{c_i\}$ are normal froms for $A$, $B$, and $AB$. In particular, $a_i|a_{i+1}$. Why must $a_i$ and $b_i$ both divide…
Potato
  • 40,171
4
votes
3 answers

Why isn't every finitely generated module free?

Suppose I have a finitely generated module $M$ over $R.$ Why is it that given a generating set I can't find a subset that is a basis? If the generating set is independent done. If it is dependent, can't we always find a maximal independent subset…
green frog
  • 3,404
4
votes
4 answers

Help understanding tensoring of exact sequences

I'm reading the proof of the following fact: Let A be a ring, a an ideal of A, M an A-module. Then $(A/a) \otimes_{A} M$ is isomorphic to M/aM. So the solution is to tensor the exact sequence $0 \rightarrow a \rightarrow A \rightarrow A/a…
user6495
  • 3,957
4
votes
1 answer

Image of a superfluous module is superfluous

Let $f$ be a homomorphism between two modules $M$ and $N$. If $K$ is superfluous in $M$ ($K$ superfluous means that if $K+L=M$ then $L=M$), then $f(K)$ is superfluous in $N$. The proof I found, but which I dont understand: Consider some $H$…
John P
  • 1,206
4
votes
3 answers

Direct sum of modules

Let $R$ be a PID,$M$ an $R$-module and $M$ is the direct sum of $M_1, \dots, M_k$ where $M_i \leq M$ for $1 \leq i \leq k$. Now let $N$ be a submodule of $M$ . Is it true that $N$ is the direct sum of $N_1,\dots,N_t$ where $N_i \lt N$ and $N_{k-t+1}…
t.k
  • 1,399
1
2
3
22 23