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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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Characterization of projective modules.

Show that $R$-module $P$ is projective, iff there exist a family $\{m_{\lambda}\}_{\lambda\in\Lambda}$ of elements of $P$ and homomorphisms $\{f_{\lambda}:P\rightarrow R \}_{\lambda\in\Lambda}$ such that $(f_\lambda(x))_{\lambda\in\Lambda}$ have…
EQJ
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Torsion and module over a ring.

I need to solve the following problem: Show that a ring $R$ is a field iff every $R$-module is a torsion-free module. The "only if" part is quite easy because if $R$ is a field then every $R$-module is a vector space then is torsion-free. I'd…
EQJ
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$N(R) \subset \operatorname{ann}_R (M)$ if M is simple left module

$N(R) \subset \operatorname{ann}_R (M)$ if M is simple left module How do I prove this? Know that if $a$ is in $N(R)$, then $a^n=0$ for some element. However, don't see how to go from this. $N(R)=\sum \{ I\}$ where I is nilpotent ideal of R.
Cohnis
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How can one show that ${\rm Hom}\Bigl(\prod\limits_{i\geqslant 1} \Bbb Z,\Bbb Z\Bigr)$ has cardinality less than $2^{\mathfrak c}$?

I have read here that it has cardinality $\aleph_0$, which follows from a theorem of Specker, which I couldn't find. I am looking for a less accurate bound to achieve the same conclusion, that $\prod\limits_{i\geqslant 1} \Bbb Z$ is not a free $\Bbb…
Pedro
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Show that $\operatorname{Hom}_R(E, F)$ is an $R$-module in a natural way.

Let $R$ be a conmutative ring, and $E,F$ are modules, show that $\operatorname{Hom}_R(E, F)$ is an $R$-module in a natural way. Is this still true if $R$ is not conmutative? I think in the to following operations: $f,g\in \operatorname{Hom}_R(E,…
EQJ
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give an example of Hom(M,N) and Hom(N,M) are not isomorphic as R modules

give an example of a commutative unitary ring R and two finitely generated R modules M and N such that Hom(M,N) and Hom(N,M) are not isomorphic as R modules.
user50276
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$\operatorname{Hom}(R\times R,M)$ isomorphic to $\operatorname{Hom}(R,M) \times\operatorname{ Hom}(R,M)$

If $M$ is a right $R$-module, then $\operatorname{Hom}(R \times R,M)$ is isomorphic to $\operatorname{Hom}(R,M) \times \operatorname{Hom}(R,M)$ My questions are: What is the isomorphism map? (I tried this $f\to(f_1,f_2)$ where $f_1(r) =f(r,0)$ and…
Mai
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Are transformations which preserve module-vectors with trivial annihilators monic?

Let $M$ be an infinite $R$-module, for a commutative ring $R$. Let $F \subseteq M$ be the set of elements which have a trivial annihilator: that is, for all $f \in F$, $$ rf = 0_M \iff r = 0_R.$$ Suppose that $T: M \to M$ preserves $F$: that is, $f…
Niel de Beaudrap
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Sections and homomorphisms of modules

Here's a problem I already solved. Prove that a homomorphism $r: M \rightarrow N$ of right A-modules admits a section $v: N \rightarrow M$ if and only if $r$ is surjective and $M = L \oplus \operatorname{ker}(r)$ where $L$ is a submodule of $M$.…
user10
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Ideal is a free module

Let $R$ be a commutative ring and consider the ideal $I = (r_{1}, r_{2})$ where $r_{1}, r_{2}$ are 2 linear independent generators of $R$. Now consider the $R-$modules. Here $I$ is a $R-$submodule of the $R-$module $R$. Can I then conclude that $I$…
postguest12
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Proving a maximal element of a certain set of ideals is prime.

Let $R$ be a commutative ring and let $M$ be a nonzero $R$-module. If $m\in M$, define $\operatorname{ord}(m) = \{r \in R \mid rm = 0\}$, and define $F = \{\operatorname{ord}(m) \mid m \in M\ \wedge m \neq 0\}$. Prove that every maximal element in…
user112127
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Let $N$ be submodule of $R$-module of $M$. Prove that if $N$ and $M/N$ satisfy DCC then so $M$ is.

Let $N$ be a submodule of $R$-module of $M$. Prove that if $N$ and $M/N$ satisfy DCC then so $M$ does. Thanks in advance.
Truong
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Proving that a module is not free.

I was reading this webpage http://ysharifi.wordpress.com/2011/10/28/examples-of-projective-modules/, and there was something that confused me. We have I thought that the dimensions of isomorphic free modules would be the same. But here it says…
user58289
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modules over group algebra which has free submodules

M and M' are modules over the group algebra $kG$ and M $\subset$ M' (submodule). Assume that M is free. Does rank(M) less or equal rank(M')? By rank(M') I mean a "minimal number of generators".
Nir Ben David
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"sum" and "products of modules (non-finite case)

Definitions: Let $A$ be a commutative and unitary ring and $M$ be an $A–$module. We will call the submodule the sum of a family of submodules $\displaystyle\{N_i\}_{i∈I}$ of $M$: $\displaystyle \sum _{i\in I}N_i=\left <\bigcup _{i\in I}N_i\right…
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