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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is (A+B)/B=A/B true for A, B modules?

If I have $A$ and $B$ two modules, is the following reasoning true? $(A+B)/B=\{a+b+B : a\in A, b \in B \}=\{a+B : a \in A \}=A/B$ I am just doubting it since it is weirdly similar to the second isomorphism theorem but still not the same thing.
roi_saumon
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Is there a name for a module that is simultaneously Hopfian and co-Hopfian?

An $R$-module $M$ is Hopfian (resp. co-Hopfian) if every surjective (resp. injective) $f:M \rightarrow M$ is an isomorphism. Do we have a name for a module $M$ that has both properties? Such a module would be a direct analogue of a finite…
Chris Leary
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Easy question on L-theory

I'm reading the introductory book on L-theory, Algebraic and geometric surgery by Andrew Ranicki, available at 1. Here is the statement in the page 254 Example 11.18. If $G$ is a group without $2$-torsion, then (there is) a decomposition $…
user8484
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What $2\mathbb Z/4\mathbb Z$ is not flat module in the ring $\mathbb Z/4\mathbb Z$

Let $R=\mathbb Z/4\mathbb Z$ be a ring and the ideal $2\mathbb Z/4\mathbb Z$ is not flat module since $2\mathbb Z/4\mathbb Z$ is generated by a nilpotent element how to prove it?
Tema
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Is every module a direct summand of a free module?

I was reading a section on Projective modules, that asked us to prove that a projective module is a direct summand of a free module. I then began thinking whether every module should be a direct summand of a free module. From what I understand, a…
user67803
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To determine the number submodules of $\mathbb{Z}_6 \times \mathbb{Z}_6$ as a $\mathbb{Z}_{12}$-module.

Determine the number submodules of $\mathbb{Z}_6 \times \mathbb{Z}_6$ as a $\mathbb{Z}_{12}$-module. Today our professor introduced the notion of modules and gave this problem as an exercise. I have thought over it and I have come up with the…
Miz
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To define a function from set {1,2} to any ring R

Let $R$ be any ring , $X$ be any non-empty set & consider $M=R^{X}$ then M is a R-module with usual addition and scalar multiplication operations Here $R^{X}=\{f\mid f:X \to R\}$ I need to use this statement to find a particular example taking $X$…
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$M=M_1\oplus M_2$. Then show that $M/M_1$ is isomorphic to $M_2$ and $M/M_2$ is isomorphic to $M_1$

Show that if $M$ is a direct sum of $M_1$ and $M_2$ then $M/M_1$ is isomorphic to $M_2$ and $M/M_2$ is isomorphic to $M_1$.
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Proof of right module over the ring $M_{2}(\mathbb{Z})$

I need to prove that $\mathbb{Z}^{2}$ is a right module over the ring $M_{2}(\mathbb{Z})$, with scalar multiplication $$(a,b)\cdot\begin{bmatrix}e & f \\g & h \end{bmatrix}=(ae+bg,af+bh)$$ I have already proved the following axiom holds: For any r…
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$M'\subseteq M$ is simple and $N=M/M'$ is simple. Then $M\cong M'\oplus N$

I am trying to solve the following problem. Let $M$ be a module over a commutative ring $A$ with identity. $M'\subseteq M$ is a simple submodule. Also, $N=M/M'$ is simple. Assume that $N$ is not isomorphic to $M'$. Prove that $M\cong M'\oplus…
learning_math
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Idea to make tensor product of two module a module structure

If M is an (S,R) bimodule and N is an left R module then How to make tensor product of M and N a S module.While making it an S module structure i got stuck in proving the property of scalar product distributive over addition.How to do that?
Tanutanu
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How to "invert a base" of a finite module in this exercise?

A is an integral domain/ring, $x \in Frac(A)$. Show that if $A[x]$ is a finite A-module, then $\exists P \in A[X]$ unitary (i.e of the form $P(X)=X^n+a_{n-1} X^{n-1}+...$) such that $P(x) = 0$ Im really stuck there. I wrote that there are elements…
Thomas
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Are finitely generated submodules of finitely generated free modules free?

Let $R$ be a commutative ring with identity. Then $R^n$ ($n$ is some positive integer) is a module over $R$ in a natural way. It is free and finitely generated. Now let $M$ be a finitely generated submodule of $R^n$. Does it follow that $M$ is free?…
Eigenfield
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Repeated application of Hom functor

For a commutative ring $R$, an $R$-module $M$ and an ideal $I\trianglelefteq R$, I want to show that $\text{Hom}_R(I, \text{Hom}_R(I,M)) = \text{Hom}_R(I^2, M)$. Is this true in general? Suppose $I$ is generated by two elements $I = (x,y)$. Then if…
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Kernel of induced map

Suppose we have $f : M\rightarrow N$ a homomorphism of $R$-modules. If $M'\leq M$, then $f$ induces a homomorphism $f' : M/M'\rightarrow N/f(M')$. What is the kernel of $f'$? Is it $(\ker f + M')/M'$? Certainly $(\ker f + M')/M' \leq \ker f'$, but…