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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$R$-algebras and the tensor-algebra.

In D & F, theorem $31(2)$ on page $442$, in the proof of the theorem given below (assuming that $M$ is an $R$-module over a commutative ring $R$ with $1$) If $A$ is any $R$-algebra and $\varphi:M \to A$ is an R-module homomorphism, then there is a…
Ben123
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Let $M$ be the $\mathbb{Z}$-module $\mathbb{Z}$. Show that the image of $\operatorname{Sym}_{2}$ is $a(1 \otimes 1)$ for even $a$.

As the title says, I want to show that Given the $\mathbb{Z}$-module $\mathbb{Z}$, we have that the image of $\operatorname{Sym}_{2}$ is $a(1 \otimes 1)$ for even $a$. My reasoning: I presume that by $\operatorname{Sym}_2$ they mean the map…
Ben123
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Submodules and summands of the right $\mathbb{Z}$-module $\mathbb{Z} \oplus \mathbb{Z}$.

Let $M_R$ be any right $R$-module ($R$ is a ring with unity). An internal direct sum $\bigoplus_{i\in I}A_i$ of submodules of $M$ is called a local summand of $M$ if for every finite subset $F\subset I$, $\bigoplus_{i\in F}A_i$ is a summand of…
Hussein Eid
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Any CS summand-square-free module is square-free

Let $R$ be a ring with unity and $M$ any right $R$-module. Recall that a submodule $B\subseteq M$ is called essential if $B\cap L \neq 0$ for every nonzero submodule $L\subseteq M$. In this case, we write $B \subseteq^{ess} M$. Recall that $M$ is…
Hussein Eid
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Where my reasoning about module homomorphisms is wrong?

I was playing around trying to find module homomorphisms and I thought I found one in the function $f:\mathbb{Z} \to \mathbb{Z}$ defined by $f(x)=2x$, as $f(x+y)=2(x+y)=2x+2y=f(x)+f(y)$ and $f(rx)=2rx=r(2x)=rf(x)$. However, if I try to apply the…
Tristan Liu
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Show that for all $m\in M$ there is $a\in A$ such that $f(m)=am$.

The problem is following: Let $M$ be a completely reducible $A$-module, $B=End_A(M)$ and let $f\in End_B(M)$. Show that for all $m\in M$ there is $a\in A$ such that $f(m)=am$. My approach: $M$ is a completely reducible $A$-module, so $M=M_1\oplus…
likeAvirgin
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Generalization of determinants to appropriate R-modules

Determinants have incredible properties in linear algebra, since to prove if a linear map $f: U \to V$, where $U,V$ have the same finite dimension, it is sufficient to show that given a basis for both spaces, and a matrix representation $B$ of $f$…
TC159
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Is there a $\mathbb{Z}$-module epimorphism $\mathbb{Z} \to \mathbb{Q}$

Consider the right $\mathbb{Z}$-modules $\mathbb{Z}_\mathbb{Z}$ and $\mathbb{Q}_\mathbb{Z}$. Is there any $\mathbb{Z}$-module epimorphism $\mathbb{Z}\to \mathbb{Q}$ ?!. Indeed, if any, it's determined only by its value at $1$. Note: By an…
Hussein Eid
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Explicit function for the Module version of the Correspondance theorem

The following I regard is the module version of the correspondence theorem Let $M$ be an $R$ module and $N \leq M$ be an $R$ - submodule of $M$. There exists a bijection $\vartheta:$ submodules of $M / N \longrightarrow$ submodules of $M$ which…
Maths Wizzard
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Let $M$ be an $A$-module, where $A$ is a commutative algebra over $R$. Is $M$ an $R$-module?

Let $R$ be a commutative ring with unity, and $A$ be a commutative algebra over $R$ with unity. Let $M$ be an $A$-module. Then can we say that $M$ is also an $R$-module? The definition of a commutative algebra over $R$ with unity that I have studied…
Saikat
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A finitely generated semisimple module satisfies the internal cancellation property

I want to prove that if $M$ is finitely generated semisimple module then it satisfies the internal cancellation property. Let $M=A\oplus B =N\oplus K$ with $A\cong N$. Since $M$ is finitely generated semisimple, then so are the summands $A,B,N$, and…
Hussein Eid
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Let $M,N$ be two submodules of an $A$-module $P$, then $(M+N)/N = N/(M\cap N)$.

This exercise is given in the lecture notes of Homological Algebra and Algebraic Topology written by Wojciech Chach´olski & Roy Skjelnes (1.8.16). Hence, there might be a typo. The sum of two submodules $M$ and $N$ is defined similarly as in vector…
Mehmet Kırtışoğlu
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$\mathbb{Z}$ as a Noetherian Module

I'm learning about Noetherian and Artinian modules. I came across an example that $\mathbb{Z}$ is a Noetherian module. My question is this; The submodules of $\mathbb{Z}$ as a $\mathbb{Z}$-module are the ideals of $\mathbb{Z}$. So stuff of the form…
sandoh
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Jacobson radical of a module equals the intersection of $IM$

Let $R$ be a unital ring and $M$ a left $R$-module. Let $\Gamma$ be the set of all maximal left ideal of $R$. I want to prove $J(M)=\cap_{I\in \Gamma}IM$. If $R$ is commutative, then $J(M)=\cap_{I\in \Gamma}IM$. proof: For every maximal submodule…
Mod.esty
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left/right adjoints of tensoring

So I have encountered the following task: Let $\phi:R\rightarrow S$ be a ringhomomorphism of commutative rings. Decide whether the functor $$\phi^*: Mod_R\rightarrow Mod_S,M\mapsto M\otimes_R S $$ admits a left/right adjoint. I know that for a given…
Adronic
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