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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A question about free module

Let R be a ring. If a direct sum of R-modules is projective, can we imply that the direct sum is also a free module over R ?
Yuan
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When the module $M$ is divisible, the radical of $M$ equals $M$.

Problem: Let $_{\mathbb{Z}}M$ be an abelian group. Prove that: If $_{\mathbb{Z}}M$ is divisible, then $\operatorname{Rad} M$$=M$. My ideas: we know that \begin{align} \operatorname{Rad}M&=\bigcap\{K\leq M\mid K \text{ is maximal in }M…
guojm
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modules over direct sum of different rings

Here, all rings are unital and $R$-modules are in fact left $R$-modules. Let $M_1$ be a $R_1$-module and $M_2$ be a $R_2$-module. Then, we can view $M_1\oplus M_2$ as a ($R_1\oplus R_2$)-module, under the operation…
valls
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$\mathbb{R}$ as a $\mathbb{Z}$ module

Is $\mathbb{R}$ free as a $\mathbb{Z}$ module? If it is, it must be of infinite rank, I suspected that some set theory will be involved, but have no clue about this fact.
mich95
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Injective linear map between modules

If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero? If not is there an obvious counterexample? (Everything is over a multivariate polynomial…
squiggles
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Definition of projective cover

In general we define the projective cover a module $M$ over an arbitrary ring $R$ as a surjective $R$-map $f: P \rightarrow M$ such that $\operatorname{ker}(f)$ is superfluous. I read (if I recall correctly in Lambek's book) that $f$ is a projective…
user6495
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Is the kernel of a map to a finitely presented module finitely generated?

Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$. Must the kernel of $f$ be finitely generated? This is true by…
User
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A little exercise about free modules

Let $R$ be a non-zero ring with unity. Let $a$ be a proper ideal of $R$. I wish to prove the following statement: $R/a$ is a free left module over $R$ iff $a=0$ where $R/a$ has the usual scalar multiplication induced by considering the left regular…
Pedro A. Matos
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What is $\ker f \otimes g$?

I am trying to find out the $\ker(f\otimes g)$ where $f:M \rightarrow P$ and $g:N \rightarrow Q$ are $A$ linear maps where $A$ is not a field. So $(f\otimes g):M\otimes N\rightarrow P\otimes Q $ is $A$ linear map.The question is: can the following…
Via
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Does projectivity over the base ring imply projectivity over the extension?

Suppose $R\to S$ is a ring homomorphism. Let $M$ be an $S$-module. Suppose $M$ is projective as an $R$-module. Is $M$ projective as an $S$-module? If not, what additional hypothesis would suffice (for example, $S$ projective over $R$)? A couple of…
user175048
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How Can I prove that this module is isomorphic to the generators with relations...

Let $G$ be a finite group of order $n$. Let $R=\mathbb{Z}G$ and $N=\sum_{g\in G} g$, observe that $gN=Ng=N$, $N^2=nN$. Let $r\in \mathbb{Z}$ be prime to $n$ and let $P_r$ be the ideal of $R$ generated by $r$ and $N$. How can I prove that $P_r$ is…
Max
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Showing that $\bigoplus_{i\in\mathbb N}\mathbb Z/2\mathbb Z$ is not a direct summand of $\prod_{i\in\mathbb N}\mathbb Z/2\mathbb Z$

I know that if the direct sum of countably many copies of $\mathbb Z/2\mathbb Z$, $I$ were a direct summand of the direct product of countably many copies, $R$ (direct summand as $R$-modules), then the complement would be a direct summand isomorphic…
Nishant
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What are the submodules of the $\mathbb{Z}$-module $\mathbb{Z}[\frac{1}{p}]$?

Given $p \in \mathbb{Z}$ a prime number and $M=\mathbb{Z}\left[\dfrac{1}{p}\right]$ a module over the ring $\mathbb{Z}$. How could I determine all submodules of M? I thougt it would work to consider the generating set $\left\lbrace 1, \dfrac{1}{p},…
Celsius
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free $\mathbb Z$-module of rank $2$

Let $M$ be a free $\mathbb{Z}$-module of rank $2$ and $p$ be a prime. Determine the number of submodules $N$ of $M$ such that $M/N \cong \mathbb{Z}/p \mathbb{Z}$. The answer may be $2$, but I cannot find a proof.
Therefore..
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Question about modules

Let $R$ be a ring and $R^n$ be the external sum of $n$ copies of the $R$-module $R$. Let's take $t$ elements of our $R$-module $R^n$ and call $M$ the sub-module generated by them. Then call $r$ the maximum number of independent elements among these…
User
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