Mathematics Stack Exchange
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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.

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How is algebra generated by a matrix defined?

I didn't find the exact definition (especially an explicit one). Let $k$ be a field, $B= \begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix} $$\in M_2(k)$ a matrix. Let $k[B]$ be the algebra generated by $B$ in…
Tom
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Equivalence between two statements in case of splitting modules

Let P be an R-module. 1.If P is a quotient of the R-module M then P is isomorphic to a direct summand of M. 2.Every short exact sequence 0→L→M→P→0 splits. How do I show equivalence between statements 1 and 2?
Sharmi C
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Isomorphism between quotient submodules

Let $A$ be a commutative ring with $1$ and let $M,N$ be $A$-modules. Let $I$ be an ideal of $A$. How do we show the following isomorphism? $(M \otimes_{A} N)/(I \cdot (M \otimes_{A} N)) \cong M/(IM) \otimes_{A/I} N/(IN)$ What confuses me is how to…
user6495
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If the Quotient $M/M'$ of a finitely generated module $M$ is a free module then $M'$ is finitely generated.

I am new to modules and I want to show that "If the Quotient $M/M'$ of a finitely generated module $M$ is a free module then $M'$ is finitely generated". Please help me.
Waqas Ali Azhar
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Proving that if M is a Right X-Module then M is also a Left Rev(X)-Module

I am very new at the topic of algebras in functional analysis and I am currently self-studying through the text "Complete Normed Algebras" by Bonsall and Duncan. In Chapter 1 section 9, page 49, the definition of a left $ A $-Module is given as…
LMW
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Is any extension ring $S \supset R$ an $R$-algebra?

Is any extension ring $S \supset R$ an $R$-algebra? In our lecture note an $R$-algebra $A$ is defined as follows $:$ An $R$-algebra $A$ is a ring $A$, which is also an $R$-module satisfying the condition $$a(xy)=(ax)y=x(ay),\ a \in R,\ x,y \in…
Arnab Chattopadhyay.
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If M a R-module then M is naturally a R/Ann(M)-Module

I am trying to show this is true by taking $(r+Ann(M)).m=r.m$. But I never use the fact that if $r\in Ann(M)$ and $x\in M$ then $rx=0$. $(r+Ann(M)).(m_0+m_1)=r.(m_0+m_1)=r.m_0+r.m_1=…
tmpys
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Can multiplication of two scalars result in zero?

In a $R$-module if $a,b\in R$ and $a\neq0, $$b \neq 0$ can it be that $ab=0$? All the modules I have seen do not allow this so I wonder if it is the case generally. The only axiom that I see which could prevent this is (when $x$ is a…
Dole
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Show that $\mathrm{End}_{R}(nM) \cong M_n(\mathrm{End}_{R}(M)) $

My attempt: Let $\psi: \mathrm{End}_{R}(nM) \rightarrow M_n(\mathrm{End}_{R}(M)) $ given by $\phi \rightarrow (\phi_{ij})$ where $\phi_{ij} = \pi_j \circ \phi \circ i_i $, where $\pi_j$ is the projection and $i_i$ is the inclusion. I need to show…
P.G
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let $M_1,..., M_n$ be left modules over R, and $M = M_1 \bigoplus ... \bigoplus M_n$.

Suppose that if $i \neq j$, then every homomorphism from $M_i$ to $M_j$ are zero. Show that $$End{}_{R}(M) \cong End{}_{R}(M_1) \times ... \times End{}_{R}(M_n)$$. Any help would be great.
P.G
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Does a finitely generated module have finitely many direct summands, up to isomorphism?

One can easily see that every finitely generated abelian group has finitely many direct summands up to isomorphism. Now, assume that $R$ is a ring and $M$ is a finitely generated $R$-module. Does $M$ have finitely many non-trivial direct…
user481657
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Let $N$, $K$ be sub-modules of $M$ with $I=\mathrm{Ann}(N)$, $J=\mathrm{Ann}(K)$. Show $I+J$ is a proper subset of $\mathrm{Ann}(N \cap K)$.

Let $N$ and $K$ be sub-modules of $M$ with $I=\operatorname{Ann}(N)$ and $J=\operatorname{Ann}(K)$. Show that $I+J$ is a proper subset of $\operatorname{Ann}(N \cap K)$.
Gargi Banerjee
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Free $R-module$ over an integral domain R that is not a field has finite length iff it is $0-module$

Free $R-module$ over an integral domain R that is not a field has finite length iff it is $0-module$ I try to begin with the most navie case:If $F$ is free $R-module$ on singleton,that is,$R$,I attempt to prove if submodule $M$ is simple,then it…
Mugenen
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If V is a non-zero completely reducible module, it contains an irreducible submodule

I'm trying to understand a proof of this from my textbook. It starts by saying that we can assume V is finitely generated, but I'm not sure why we can make that assumption. Is there something that goes wrong in the infinitely generated case? Other…
Joe
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Submodule of a finitely generated module over an artinian ring

I have a left artinian ring $R$ and a finitely generated left $R$-module $M$, and a submodule $A$ of $M$. My question is : is $A$ necesseraly finitely generated ? (and is there a direct proof of this from the "artinianity" of $R$ ?)
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