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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Is the $k$-algebra $k[G]$ semisimple?

Let $k$ be an algebraically closed field of characteristic $p$, and $G$ a cyclic group of order $p$. I was asked to show that $k[G]$ is not semisimple. But I get confused since for any $k_1g_1 + k_2 g_2 \in k[G]$, $ (k_1g_1 + k_2 g_2)^p = k_1^p…
user112564
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Are two bimodules isomomorphic as left and right modules also isomorphic as bimodules?

let R be a commutative ring, and M, N two bimodules over R, such that there exists f : M -> N an isomorphism of left R-modules, and g : M -> N an isomorphism of right R-modules. Then are M and N isomorphic as R-bimodules? I don't see any reason why…
FrenchStudent
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Every $A$-mod is flat iff every mod-$A$ is flat.

$\require{AMScd}$ Suppose $A$ is a ring, and we have a SES of left $A$-modules $$0\longrightarrow M'\mathop\longrightarrow\limits^f M \mathop\longrightarrow \limits^g M''\longrightarrow 0$$ where $M''$ is flat. Now let $N$ be any right $A$-module.…
Pedro
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Suspicion about simple modules

It's stated here that for a commutative ring $R$, every simple module over $R$ is isomorphic as an $R$-module to a quotient ring of $R$ by a maximal ideal. Intuitively this seems likely, since a quotient ring is a field if any only if we quotient by…
Ashley Lin
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Free modules are generators

Let $A$ be a ring. We say an $A$-module $M$ is a generator if for every $A$-module $N$ there is a surjective $A$-homomorphism $M^{(I)} \rightarrow N$ where $M^{(I)}$ denotes the $|I|$ copies of the direct sum of $M$. Why is it true that every free…
user10
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The module is square-free iff every factor is summand-square-free

Recall that a right $R$-module $M_R$ is called square-free if whenever $A,B\leq M$ are submodules with $A\cap B=0$ and $A\cong B=0$ then $A=B=0$. A submodule $N$ of a module $M$ is called a summand (written $N \leq^{\oplus}M$) if $M$ is the internal…
Hussein Eid
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Basis for a $R$-Module $R$, with $R$ the ring of endomorphisms

I got a doubt with this problem: Let $M=\{f:\mathbb{N}\to \mathbb{Z}|\text{$f$ is a function}\}$, defining the sum in $M$ as $(f+g)(n)=f(n)+g(n)$, $M$ is an abelian group. Let $R=\{\phi :M\to M : \phi\ \text{is a morphism}\}$. $R$ is a ring with…
Juan Pablo
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Prove that $\eta: R/(a)\to R/(p_1^{\alpha_1})\oplus \cdots \oplus R/(p_r^{\alpha_r})$ is an epimorphism

Let $R$ be PID, and $a \in R$ with $a=p_1^{\alpha_1}\cdots p_r^{\alpha_r}$ its prime factorization. I'd like a Hint in proving that $$ x+(a) \to \eta(x+(a))=(x+(p_1^{\alpha_1}),\ldots,x+(p_r^{\alpha_r})) $$ is an epimorphism. My approach is too…
Juan Pablo
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Direct sum problem in module theory

Show that $\mathbb{Z}/(p^n),p$ a prime $n\ge 0$, regarded as $\mathbb{Z}$-module is not a direct sum of any two non-zero submodules. Does this hold for $\mathbb{Z}/(n)$ for other positive nteger ??? Please help anyone .
Aritra Roy
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A question about R-modules.

Let $M$ be a finitely generated R-Module; $m\in M$. $b_{1}, b_{2}\ldots b_{n}$ is the minimal generating set of $M$. I read somewhere that if $m=s_{1}b_{1}+s_{2}b_{2}\ldots s_{q}b_{q}=s_{1}'b_{1}+s_{2}'b_{2}\ldots s_{q}'b_{q}$, then it is not…
user67803
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Modules isomorphic over subring

Let $R$ be a commutative ring with identity. Let $M,N$ be two isomorphic $R$-modules, and let $S \subset R$ be a subring of $R$, sharing a common identity element. Is it then true that $M$ and $N$ are isomorphic as $S$-modules? My intuition says…
Trettman
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If $M$ is a $R$ Module which has a finite generating system and $T$ is a submodule of $M$, is there a submodule $F$ such that $M=F\oplus T$?

We know that in vector spaces such a complementary substructure exists. But in this proof we took advantage of the fact that every vector space has a basis, ie. the subvectorspace has a basis and this basis then can be extended to a full basis of…
New2Math
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Examples of modules such that $M\oplus N_1\simeq M\oplus N_2$, but $N_1\not\simeq N_2$?

Are there any examples of modules such that $M\oplus N_1\simeq M\oplus N_2$, but $N_1\not\simeq N_2$ as modules? I thought of taking $\bigoplus_{i\geq 0}\mathbb{Z}\oplus\mathbb{Z}\simeq\bigoplus_{i\geq 0}\mathbb{Z}\oplus\{0\}$ as…
user62738
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About the inductive step on the Structure Theorem on Principal Domains

Im looking for a proof on the following statement Let R be a principal domain and M be an R-module of p-torsion and finitely generated ($p\in R$ prime). Let $x_1 \in M$ with $Ann_R(M)=Ann_R(x_1)$, and consider the quotient $N=M/R.x_1$. If…
Marcos Martínez Wagner
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Endormorphism rings over simple modules

I know that if $U$ is a simple left $R-$module then $\text{End}_R(U)$ is a division ring. It is an easy argument. But I'm struggling to see what happens when we move up to trying to classify endomorphism rings of semisimple modules. Even worse, I…
J. Moeller
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