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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Structure of $k$ as $k[X]$-module

Let $k$ be a field. There are several ways to make $k$ into a $k[X]$-module; for example, by fixing $\alpha \in k$ and taking $P(X) \cdot t := P(\alpha)t$ for all $t \in k$. Are there other obvious ways to make $k$ into a $k[X]$-module? Can we…
Evariste
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Why does this module isomorphism hold?

I'm a little unsure of why the following module isomorphism holds. Say $R$ is a commutative, unital ring and $\mathcal{M}$ an $R$-module. If $I_1,\dots,I_n$ are pairwise comaximal ideals of $R$, then $$ M/(I_1\cdots I_n)M\cong\bigoplus_{i=1}^n…
Danielle Intal
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Is the $k$-algebra of upper triangular matrices an indecomposable algebra?

Let $k$ be an algebraically closed field. Let $U_{n}(k)$ denote the $k$-algebra of upper triangular matrices of size $n \times n$ over $k$. Is it true that this $k$-algebra is indecomposable? Is this because its socle is isomorphic to $k^{n}$ and…
user10
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Module and direct sum

$R$ is a ring and $I$ is a left ideal. If $R/I$ is an $R$ module and projective. Does this mean $R/I \oplus M \cong R$ for some $R-$module $M$? I know that $R/I \oplus M \cong R^{n}$ for some $n$ and $M$.
postguest12
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A module trapped between two free modules is finitely generated

Let $R$ be a Noetherian integral domain, $M$ and $R$-module and $F_1$, $F_2$ free $R$-modules of rank $n$ for which $$F_1 \subseteq M \subseteq F_2$$ Is it true that $M$ is finitely generated? If so, is there a "slick" proof of this (perhaps using…
user114722
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If $R$ is a ring that is finitely generated as an additive group, then R is Noetherian

Recall that a ring $R$ is called right (left) Noetherian if every right (left) ideal $I$ of $R$ is a finitely generated $R$-module, i.e., there exists $x_1,\ldots,x_m \in I$ such that $I=x_1R+\ldots+x_mR$ (or $I=Rx_1+\ldots+Rx_m$). Suppose that $R$…
Hussein Eid
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$A$-algebra B and $Hom_{A}(B,A)$ and their rank.

Let $A$ and $B$ be a commutative ring and $B$ an $A$-module, and suppose that $B$ is finitely generated and free as an $A$-module. Is $Hom_{A}(B,A)$ free over $A$ with the same rank as B? Thanks in advance!
user53216
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Any artinian module is a finite direct sum of indecomposable modules

Let $M_R$ be an artinian module. I proved that there exist $n\in \mathbb{N}$ and nonzero uniform submodules $U_1,\ldots,U_n$ of $M$ such that $U_1+\ldots+U_n$ is a direct sum and $\bigoplus_{i=1}^n U_i$ is an essential submodule of $M$. In fact, the…
Hussein Eid
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Is there a non-negative integer $n(M)$ for a module $M$ with these conditions?

Let $R$ be a ring and $M$ be an $R$-module. Also assume that $N$ is a complement in $M$ (recall that a submodule $K$ of $M$ is called a complement in $M$ if there exists a submodule $L$ of $M$ such that $K\cap L=\{ 0\}$ and $M=K+L$. In other words,…
M.Ramana
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If $K \subseteq^{\mathrm{ess}} N \subseteq M$ implies $N/K$ is f.cog for all $N,K$ , then is $M/A$ f.cog for any $A \supseteq \mathrm{soc}(M)$?

Let $R$ be a ring with unity and $M$ any unitary right $R$-module. We say that $M$ is finitely cogenerated if whenever $\lbrace A_\lambda \rbrace_{\lambda\in \Lambda}$ is a collection of submodules of $M$ with $\bigcap_{\lambda\in \Lambda}…
Hussein Eid
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Equivalence of two definitions of module length

Usually, we say a module is of finite length if it has a composition series. We then use the composition series to define its length (after showing the Jordan Holder theorem). I'm trying to work with an alternative definition and show that they are…
suncup224
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An idempotent Ring-module homomorphism

I want to prove that if $f:A \rightarrow A$ is an R-module homomorphism such that $f\circ f=f $ then $A= Ker f\oplus Im f $ If we think about a split exact sequence like : $ 0\rightarrow Ker f \rightarrow A \rightarrow Im f \rightarrow 0 $ then we…
NotaChoice
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Is the sum of a family of simple submodules always interal direct sum?

Let $\lbrace S_i \rbrace_{i\in I}$ be a family of pairwisw disjoint non-isomorphic simple submodules of a module $M_R$. Is it true that the sum $\sum_{i\in I}S_i$ is internal direct sum (that is, $S_k \cap \sum_{i\neq k}S_k =0$). If so, how could I…
Hussein Eid
  • 1,049
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$\mathbb{Q}$ is a square-free $\mathbb{Z}$-module

A module $M$ is said to be a square-free module if for any two submodules $A,B \leq M$ with $A \cap B = 0$ and $A \cong B$ we have that $A=B=0$. I claim that $\mathbb{Q}$ is a square-free as a left $\mathbb{Z}$-module. Does anybody know how to prove…
Hussein Eid
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Uniserial module definition

Let $M$ be an $R$-module, $M$ is said to be uniserial if submodules of $M$ are totally ordered by inclusion. That is if $N$ and $L$ are two submodules of $M$ the either $N\subseteq L$ or $L\subseteq N$. My question is can we drop equality here like…
sushil soni
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