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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Difference between free and finitely generated modules

I am not sure I understand the difference between free modules and finitely generated modules. I know that a free module is a module with a basis, and that a finitely generated module has a finite set of generating elements (ie any element of the…
user62406
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Every module is the quotient of a free module.

Is every module a quotient of a free module by the following? Suppose I have a module $M.$ Take the free module $F = \oplus_{m \in M}R.$ Construct a surjective module homomorphism by sending the element with 1 at the $m^{th}$ coordinate and 0…
green frog
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A module without a basis

It is well known that every vector space have a basis, after studying module theory I read in Wikipidia that not all modules have a basis. Can someone please give an example of a module without a basis ?
Belgi
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Are submodules of finitely generated free modules always free?

I don't believe it's true that all submodules of a finitely generated free module are free just based on results from google. Is there a canonical simple example of a finitely generated free module with a submodule that is not free?
Garry
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Module isomorphic to second dual

Is there a simple condition on a module $M$ over a ring $R$ which will ensure that $M$ is isomorphic to its double dual, $M^{**} = \operatorname{Hom} (\operatorname{Hom}(M,R),R)$? What about a condition on $R$ which guarantees this will hold for all…
user15464
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Why is this a $k$-module?

Just started studying tensor products. Let $A$ be a commutative ring with unity and let $M$ be an $A$-module. Now let $k$ be a field, I know that a $k$-module is precisely a $k$-vector space. My question is the following: Why $k \otimes_{A} M$ is…
user8632
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Module over a quotient ring

During my revision, I came across this exercise: Let $R$ be a ring and $I$ an ideal of $R$. If $M$ is a simple left $R/I$-module, then $M$ is a simple left $R$-modules with operation define as follows: $rm=(r+I)m$, for all $r \in R$ and $m \in…
user39280
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restriction of scalars, reference or suggestion for proof

Let $f:R \rightarrow R'$ be a ring homomorphism that is epic in the category of rings. Let $M,N$ be $R'$-modules. Why is it that a homorphism $h:M \rightarrow N$ is $R'$-linear if and only if it is $R$-linear? I have used this result in specific…
Chris Leary
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Prove that $M \otimes N$ is isomorphic to $N \otimes M$.

Let $M,N$ be left $R$-modules with the standard $R$-module structures (making them bi-modules). The proof that $M \otimes N$ is isomorphic to $N \otimes M$ over a commutative ring $R$ in Dummit and Foote says that, 1) the map $(m,n) \mapsto n…
Jean Valjean
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Does restriction of scalars preserve projectivity?

Suppose $A \to B$ is a ring homomorphism. If $P$ is a projective $B$-module, is $P$ projective as an $A$-module? I can prove this with the added hypothesis that $B$ is projective as an $A$-module, but is it true more generally?
user74897
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Help understand isomorphism of tensor product and connection to vector spaces

well I'm having a hard time understanding the tensor product. Here's a problem from Atiyah and Macdonald's book: Let $A$ be a non-trivial ring and let $m,n$ be positive integers. Let $f: A^{n} \rightarrow A^{m}$ be an isomorphism of $A$-modules.…
user6495
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Showing $\mathbb{Z}_6$ is an injective module over itself

I want to show that $\mathbb{Z_{6}}$ is an injective module over itself. I was thinking in using Baer's criterion but not sure how to apply it. So it suffices to look at non-trivial ideals, the non-trivial ideals of $\mathbb{Z_{6}}$ are: (1)…
user10
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Why is there only one way to make $M$ a $\mathbb{Q}$-module, if possible?

Suppose $M$ is a $\mathbb{Q}$-module. Why is the given action of $\mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $\mathbb{Q}$-module? I know that an abelian group can only be made into a $\mathbb{Z}$-module in a unique way, since…
Cellarius
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About the definition of a module.

The following is the definition of a module from the Wikipedia. Suppose that $R$ is a ring and $1R$ is its multiplicative identity. A left $R$-module $M$ consists of an abelian group $(M, +)$ and an operation $R × M → M$ such that for all $r, s$ in…
user112018
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Cancellation law in finitely generated modules

Let $R$ be a commutative ring with $1$ and let $S,T$ be $R$-modules such that $S$ is finitely generated and such that $S \cong S \oplus T$. Must $T=0$? This is certainly true if $R$ is a PID, but what if $R$ is just a commutative ring with $1$? (if…
user6495
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