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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Every right ideal of R is injective

How can I prove the following statement: Every right ideal of $R$ is injective iff $R$ is semisimple. It's a strange statement. If true, only from the condition satisfied by the ideals we can conclude a property valid for all modules over the ring.
Vasco
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Indecomposable, infinitely generated, non-isomorphic modules over the integers

I am having trouble proving the following statement. Let $A$ and $B$ be infinite, disjoint sets of prime numbers with $5 \notin A\cup B$. Set $e_2' = 5e_1+2e_2$ where $e_1,e_2,e_3$ denotes the canonical basis of $\mathbb{Q}^3$. Consider the two…
Anders G.
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Do all submodules of $R^n$ have a complement in $R^n$?

Suppose that $M=R^n$, where $n\geq 1$ and $R$ is a PID and suppose $N\leq M$. A \emph{complement} of $N$ in $M$ is a submodule $P$ of $M$ so that $M=N\oplus P$ (internal). If $A\in M_{n\times n}(R)$, the \emph{nullspace} of $A$…
Boga
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Problem in the proof of whether two bases are equipotent or not of an $R$ module $M$

I am reading T.S.Blyth's Module Theory where it is written that two bases of an $R-$ module may have different cardinalities but if the ring $R$ is commutative then any two bases(if exist) of an $R$ module $M$ are equipotent. In the proof it uses a…
Learnmore
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Property of free modules.

Let $R$ be a PID. I want to show that if $P$ is a finitely generated left $R$-module and $P$ is isomorphic to $R^n$ and $R^m$ (as $R$-modules) for $n,m$ in the natural numbers, then $n=m$. I was wondering if the following proof is enough: Assume…
Emma
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Can one extent to (left) modules on IBN ring the results of vector spaces?

We know that (left) modules "are not as nice as vector spaces" because lots of properties of these ones don't fit to the most general modules. But il we restrict to modules on IBN ring (i.e. modules in which rank is well-defined), are we "at home"…
Carré rond
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$_RM_S$ is R-projective and $_SN$ is S-projective then $_RM \otimes_S N$ is R-projective.

I want to prove that if $_RM_S$ is R-projective and $_SN$ is S-projective then $_RM \otimes_S N$ is R-projective. Projectivity of $_RM$ and $_SN \implies M \oplus K \cong R^{(I)}$ and $N \oplus L \cong S^{(J)}$ My question is: I want to construct…
Luis Vera
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Different definitions of submodule

I have stumbled upon two different definitions of a submodule and I cant see why they are equivalent, let me state them ($M$ is a module and $R$ a commutative ring): First Definition: A set $N \subset M$ is called a submodule of $M$ if $$an + bm \in…
jdaUU
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About submitting a paper in algebra?

My questions are about submitting papers related to modules category: (1) how much does it take (usually) to get the acceptance from a journal with good impact factor? (2) which journals(with good impact factor) take less time? (3) can I submit my…
Sms
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If $R$ is a commutative ring then prove that every left $R$-module is also a right $R$-module

$R$ is a commutative ring and $M$ is a left $R$ module. Define $\cdot : M*R \to M$ as $(x,a) \to x\cdot a$ by, $$x\cdot a=ax$$ To prove : $M$ is a right $R$ module. Proof: Let $x,y \in M$ and $a,b\in R$, Then the first two properties i.e.…
johny
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If $\mathfrak{m}M = o$, then the $R$-module $M$ is equivalently an $R/\mathfrak{m}$-module.

While I was reading a book, I encountered this proposition: Suppose that $\mathfrak{m}$ is an ideal of a ring $R$ and that $M$ is an $R$-module. If $\mathfrak{m}M = o$, then the $R$-module $M$ is equivalently an $R/\mathfrak{m}$-module. I just…
Behrooz
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Finding Modules with Given Length of Composition Series

I study a course about commutative algebra, and I saw many questions as the following: Find an example of a $\mathbb{Q}\left[\lambda_1,\lambda_2\right]$-module with $\ell\left(M\right)=3$ (i.e., $M$ has a composition series of length 3. Can you…
Guy
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a question about artinian modules

Let M be a modules. Are the following equivalent. i) M is artinian ii) every factor of $M$ is finitely generated. Yours,
Takj
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$f:M\to N$ is surjective $R$-linear map

$R$ be a commutative ring with $1$ and maximal ideal $m$. Let $f:M \rightarrow N$ be an $R$-linear map and $N$ is a finitely generated $R$-module. Suppose that the induced homomorphism $M\otimes_R(R/m) \rightarrow N\otimes(R/m)$ is surjective. Now…
Via
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Semisimple submodule

Suppose that $M$ is a semisimple module, i.e $M \cong \oplus_{i \in I} S_{i}$ where $S_{i}$ are simple modules. Let $N$ be a submodule of $M$. Why there exists a subset $J \subseteq I$ such that $N \cong \oplus_{j \in J} S_{j}$?
user10
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