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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Does a homomorphism $x^k\mapsto k$ exist?

If $M=\mathbf{Z}[G]$ is the group considered as a module over itself, and with $\mathbf{Z}$-basis $\{1,x,\ldots,x^{n-1}\}$, is it possible to construct a module homomorphism $\varphi:M\to\mathbf{Z}$ such that $\varphi(x^k)=k$? If so, how do we show…
Sam Williams
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tensor an exact sequence

Let $R$ be a ring, $M$ an $R$-module, and $I$ an ideal of $R$. Then given the exact sequence: $$0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow0$$ when is $M$ flat. By some theorem, $M$ is flat if and only if $I\otimes…
claire
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Definition of direct summand

If M is an R-module ,and if N is a direct sum of M over an indexed set I, then is it correct to write: N= T direct sum M, where T is the direct sum of M over I-{i} , where i in I.
Until
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Some questions about the properties of submodule

I am reading a book on Abstract and Linear Algebra and saw the following theorem: "Suppose $M$ is an $R$-module, $T$ is an index set, and for each $t \in T$, $N_t$ is a submodule of $M$. i) $\cap_{t\in T} N_t$ is a submodule of $M$. ii) If {$N_t$}…
velut luna
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When is an element of this tensor product equal to $0$?

Let $A$ be an abelian group (i.e., a $\mathbb Z$-module), and consider the tensor product $\mathbb Q\otimes_{\mathbb Z} A$. I want to show that the element $\frac{1}{d}\otimes n=0$ iff there's a positive integer $r$ such that $rn=0$. I have shown…
Nishant
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example of tensor product of three finitely generated R-modules that equal 0

Can you give an example of a commutative unitary ring $R$ and three finitely generated $R$-modules $L$, $M$ and $N$ such that $L\otimes M\ne 0$, $M\otimes N\ne 0$, and $L\otimes N\ne 0$, but $L\otimes N\otimes N = 0$?
user50276
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Extension of R-modules.

Let $$ 0\longrightarrow{A}\longrightarrow{B}\longrightarrow{C}\longrightarrow{0} $$ be a short exact sequence of $R$-modules $A,B,C$ then why do we call $B$ is an extension of $C$ by $A$?
user54992
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Showing that finitely generated module forms a torsion quotient.

I have that $R$ is a domain and $M$ is a finitely-generated $R$-module. I am trying to show that there is a free submodule $F$ such that $M/F$ is a torsion $R$-module. I have the fact: If $X$ is a torsion generating set of $M$ then $M$ is…
Wooster
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Projective module, polynomial ring

Let $k$ be a field and consider $k[x,y]$. $k$ is then a $k[x,y]-$module if we let $k[x,y]$ act by $f(x,y)\cdot k = f(0,0)\cdot k$. Is $k$ projective? I have tried to construct an isomorphism between $k\oplus (x,y) \cong k[x,y]$
postguest12
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Basis for a module.

Clearly $\mathbb{Z}$ and $\mathbb{R}$ are $\mathbb{Z}$-module. Then is $2\mathbb{Z}$ a submodule of $\mathbb{Z}$? What is a basis? And what is a basis for $\mathbb{R}$?
user112018
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Classification of $M_{n}(\mathbb{Z})$-submodules of $\mathbb{Z}^n$

I recently saw this question and wanted to generalize it to $\mathbb{Z}^n$ and matrices with integer coefficients. I think $k\mathbb{Z}^n$ is a submodule for any integer $k$. Is there a nice way to generalize this?
Boran Erol
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How to show :/⊗r→/ is a well-defined R/I module homomorphism

I'm new to tensor products, and am struggling with showing that f is a well-defined R/I homomorphism, where I is an ideal of R ring, and A is a left R module. Showing scalar multiplication isn't too bad , as f((x+I)(r+I X a)) = f(xr + I X a) = xra +…
John Li
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Show that if $M$ is an $R$-module that contains a free submodule $N$ of rank $n$ such that $M/N$ is torsion, then $M$ is of rank $n$.

Assume that $R$ is and integral domain. If $M$ is an $R$-module that contains a submodule $N$ free of rank $n$, such that $M/N$ is a torsion $R$-module, then $M$ is of rank $n$. My reasoning: That $M/N$ is a torsion-module implies that for each…
Ben123
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Unique $R$-algebra homomorphism from the exterior algebra to an $R$-algebra $A$, if $a^2 = 0$ for all $a \in A$.

Let $A$ be an $R$-algebra for a unital commutative ring $R$, such that $a^2 = 0$ for all $a \in A$ and $\varphi:M \to A$ be an $R$-module homomorphism. We want to prove that it follows that there exists a unique $R$-algebra homomorphism…
Ben123
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Right and Left modules.

Let $R$ be a ring and M an abelian group such that: (i)$(r+s)m=rm+sm$, $r,s \in R$, $m \in M$ (ii)$r(m+n)=rm+rn,$ $r \in R, m, n \in M$ (iii) $r(sm) = (rs)m,$ $r,s \in R$, $m \in M$. Show that $m \cdot r := rm$ defines a right $R$-module on $M$.…
user58289
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