Mathematics Stack Exchange
Mathematics Stack Exchange

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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$A$: ring, $S$ multi closed subset. $M:A$-module $\iota:M\to S^{-1}M, m\mapsto m/1$. $N'$: $S^{-1}A$-submodule, $N=\iota^{-1}N'$. Then $S^{-1}N=N'$

Let $A$ be a ring and $S\subset A$ be a multiplicatively closed subset. Let $M$ be an $A$-module and let $\iota:M\to S^{-1}M$ be the natural map given by $m\mapsto m/1$. let $N'$ be a $S^{-1}A$-submodule of $S^{-1}M$, and let $N:=\iota^{-1}N'$ be…
Edelweiss Ntu
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What is the rank of $\Bbb{Z}_m\oplus\Bbb{Z}_n$?

At first glance, the rank seems to be $2$. The basis elements seem to be $(0,1), (1,0)$. However, how is scalar multiplication defined? Is $a(p,q)=(ap,q)=(p,aq)$? So if the module under consideration is $\Bbb{Z}_2\oplus \Bbb{Z}_3$, is…
freebird
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two sided ideals and idempontents

Let I be a two sided ideal of R. Prove that I=eR for some central idempotent e ϵ R if and only if R=I+J for some two sided ideal J. When this occurs, show that e and J are uniquely determined by I. Attempt: e is a central idempotent of R then e^2=e…
Uthando
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given a surjective ring homoromorphism from R to S, ideals in S are the R submodules of S

A lemma in my lecture notes claims, as in the title, given a surjective ring homoromorphism from R to S, ideals in S are the R submodules of S (in the first paragraph - picture attached), but I can't see why this is true. Note that we are assuming…
lkjhgfdsa
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When does isomorphism between submodules induce an isomorphism between modules

Let $R$ be a ring and let $M$ and $N$ be $R$-modules with submodules $L$ and $P$, respectively. Assume we know $M \cong N$ as $R$-modules. Further, suppose we are given an isomorphism $$\psi: L \to P$$ When is it true that there exists an…
Exit path
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Prove that $\operatorname{Ann}_{R}(S)$ is a two-sided ideal of $R$ whenever $S$ is a submodule of $R$-module $M$

Let $M$ be an $R$-module and $S$ be a subset of $M$. The annihilator of $S$ in $R$ is $\operatorname{Ann}_{R}(S) = \{ \lambda \in R; \forall x \in S \ \ \ \lambda x = 0 \}$. I need to show that $\operatorname{Ann}_{R}(S)$ is an ideal(two-sided) of…
Jxt921
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Isomorphism as $R$-module

Let $R = M_2(\mathbb{C})\times M_2(\mathbb{C})$ and let \begin{equation*} e_1 = \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right),~~ e_2 = \left( \begin{bmatrix} …
Learner
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Finitely Co-generator

Let M be an R-module. We say that M to be Co-generator. If for any family $(M_{i})_{i\in I}$, I is non-empty set, $M_{i}$ is submodule of M such that ${O_{M}}=\cap_{i\in I} M_{i}$ then there is $B\subset I$ is finite that ${O_{M}}=\cap_{i \in B}…
PHU CUONG LE VAN
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Why is the direct product of modules not a module?

My book says the following: Let $M $ and $N $ be modules. Then $M\times N $ is not a module. I don't understand this statement. It seems to satisfy all properties of modules.
user67803
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$\phi(X)$ is essential and $\psi \circ \phi$ is injective then $\psi$ is injective.

Let $R$ be a commutative ring, and $M$ an $R$-module. An $R$-submodule $N$ of $M$ is said to be essential if for any $R$-submodule $U$, $U \cap N = \{ 0 \}$ then $U = \{ 0 \}$. My problem is the following: Let $\phi:X \rightarrow Y, \psi:Y…
user
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Question on image of functor.

Let $R=\mathbb{Z}[x]$. Define $F: \text{Mod}_R \to \text{Ab}$ by $F(M)= \{x \cdot m : m \in M \} \subseteq M$, it can be seen that $F$ defines a covariant functor. Next, I must check that the functor is exact, but in order to talk about exact…
Hujiik
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If $M$and $N$ are R- modules, then under what conditions $\operatorname{Hom}(M,N)$ the space of R-module morphisms from M to N, is projective?

If $M$ and $N$ are $R$- modules, then under what conditions $\operatorname{Hom}(M,N)$ is projective? I was trying to show that $\operatorname{Hom}(M,N)$ might be written as tensor product of two modules, i.e. Of dual of $M$ and $N$ (like in case of…
satyendra
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About extensions of finitely generated modules.

Let $R$ be a ring with unity, and $A \subseteq B$ be $R$-modules. I want to know if the following is true: If $A, B/A$ are finitely generated $R$-modules, then $B$ is finitely generated. Under some stronger hypothesis, this is true. For…
Crostul
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Show that any two elements of $M$ are linearly dependent over $\mathbb Z$.

Let $M$ be the additive group of rationals .Show that any two elements of $M$ are linearly dependent over $\mathbb Z$. I do not understand why is this true.If $c_1\dfrac{a}{b}+c_2\dfrac{c}{d}=0$ then $c_1=c_2=0$ .Is this result true or a fault in my…
Learnmore
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Module over different rings

If $M$ is a finitely generated abelian group, which can be made into a module over a ring $R$ (with a certain scalar multiplication) and has as a module the minimal spanning set $\{e_1, \ldots, e_n\}$, is it possible that a different ring $S$…
Ystar
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