For questions about modules over rings, concerning either their properties in general or regarding specific cases.

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2answers
81 views

Who is $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}^m,\mathbb{Z}^n)$ isomorphic to?

The problem is: Who is $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}^m,\mathbb{Z}^n)$ isomorphic to? (where ...
2
votes
0answers
69 views

Proof that $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a finitely-generated left $A_1(\mathbb{C})$-module

I want to prove that $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a finitely-generated left $A_1(\mathbb{C})$-module where $\lambda \in \mathbb{C}$ and $\mathbb{C}[x,(x-\lambda)^{-1}]$ is a subring of ...
3
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1answer
85 views

Examples of non-principal free ideals

If $R$ is a commutative ring, and $I\subset R$ is a non-zero free ideal, then it is principal generated by a non-zerodivisor. If $R$ is a non-commutative ring having the IBN property and $I\subset R$ ...
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1answer
106 views

Example of a ring such that $R^2\simeq R^3$, but $R\not\simeq R^2$ (as $R$-modules)

The usual example of unitary ring without the IBN property is the ring of column finite matrices, and in this case we have $R\simeq R^2$ as (left) $R$-modules. (See also here.) In particular, we have ...
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1answer
52 views

Example of an $R$-module $M$ that is not simple but $\mathrm{End}_R(M)$ is a field.

I came across Schur's theorem: If an $R$-module $M$ is simple, then $\mathrm{End}_R(M)$ is a skew field. ($M$ is simple if the only submodule are $M$ and $0$, and $M \neq 0$.) I heard that it ...
2
votes
1answer
60 views

Induction functor from $\mathbb{K}$-mod to ($\mathbb{K}\times\mathbb{K}$)-mod

Let $\mathbb{K}$ a field. Given a $\mathbb{K}$-mod (a vector space) there is an induction functor on $\mathbb{K} \times \mathbb{K}$-mod that is, as usual, $- \otimes_\mathbb{K} (\mathbb{K} \times ...
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1answer
13 views

Intersection module

Let M be an R-module, $M_{1}, M_{2}$, $K$ be three submodules such that $K \subseteq M_{1}, K \subseteq M_{2}$. The question is: $\dfrac{M_1}K\cap\dfrac{M_2}K=\dfrac{M_1\cap M_2}K$ ? where ...
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1answer
14 views

A problem involving proving cyclic module defined via invertible linear operator has cyclic inverse module

I have met this recently in my abstract algebra course dealing with modules over PIDs and we are dealing with cyclic modules at the moment, the problem I am having difficulty with is as follows: ...
4
votes
1answer
35 views

Justification for thinking of modules over commutative rings as modules of generalized functions?

Given a commutative ring one may look at it geometrically in terms of its affine scheme - a locally ringed space associated to it, on which its elements behave similarly to functions. I've read here ...
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1answer
48 views

How is an abelian $G$-operator group with $m1 = m$ a $\mathbb Z[G]$-module

Let $M$ and $G$ be groups. We call $M$ a $G$-group (or group with operators) if every $g \in G$ corresponds to an endomorphism of $M$, i.e. we have $$ (mn)^g = (n^g)(m^g). $$ (the application of ...
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2answers
29 views

socle(M) being simple gives an upper bound for the dimension of End(M)?

Suppose $k$ is a algebraically closed field of arbitrary characteristic. Let $A$ be a finite dimensional $k$-algebra and $M$ an $A$-module with finite dimension with respect to $k$. I have seen it ...
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0answers
15 views

Counter example that Artinian k-algebra is not finite k-vector space [duplicate]

Let k be a field, A be a k-algebra. If A is not a finitely generated k-algebra,then the following two conditions are NOT equivalent: (i) A is Artinian; (ii) A is a finitely k-algebra, i.e. A is ...
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0answers
12 views

Computing the invariant factors and elementary divisors of finitely generated module over a PID

I have just encountered this question in my abstract algebra class dealing with finitely generated modules over PIDs stating the following: Let $ D = \mathbb{R}[x] $ be the ring of polynomials ...
4
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1answer
58 views

How are $G$-modules and linear group actions different

Let $M$ be an abelian group and let $G$ be a group acting on $M$ such that $M$ is a $G$-operator group, i.e. we have for $u, v \in M$ and $g,h \in G$ (1) $u\cdot 1_G = u$ (2) $(ug)h = u(gh)$ (3) ...
2
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1answer
30 views

Shift of a simple graded module

I am trying to understand the simple graded modules over a graded ring $R$ (all the gradings over $\mathbb{Z}$). I know that there exists a bijection between the simple graded $R$-modules and simple ...
3
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2answers
84 views

How to show that $\mathbb Q/\mathbb Z$ is not a $\mathbb Q-$module?

How to show that $\mathbb Q/\mathbb Z$ is not a $\mathbb Q-$module ? I tried to define an action $\mathbb Q\times \mathbb Q/\mathbb Z\longrightarrow \mathbb Q/\mathbb Z$ but I can't conclude.
4
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2answers
52 views

If $f: A\to B$ is faithfully flat and $B$ is an Artinian ring then $A$ is also Artinian.

Let $f : A → B$ be a map of rings. The map $f$ is called faithfully flat if $B$ is flat $A$-module ($B$ is $A$-module w.r.t. multiplication defined by $ab := f(a)b$) and if for any $A$-module $M, M ...
1
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1answer
33 views

Step in proof that a quotient is isomorphic to cohomology group

The following proof is about the cohomology group. But the fact I do not understand is something like, we have two groups $U \cong V$ and two normal subgroups $N \cong M$ in $U$ respectively $V$, what ...
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0answers
21 views

Prerequisites for Blyth's Module theory - an approach to linear algebra

I would like to know from people who have read this book if it can be tackled without prior exposure to linear algebra. The author claims in the introduction that "algebraic prerequisiste is the ...
3
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2answers
49 views

Modules over associative algebras are just special cases of “ordinary” modules over rings?

By module over a ring, I mean always a right-module. All rings are supposed to be unital, and the module fulfills $m\cdot 1 = m$. If $R$ is commutative and $M$ a right-module, we can define $rx := xr$ ...
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0answers
34 views

Prove that for a domain $A$, $A^n$ cannot have $A^m$ as a subring where $m>n$

I have to prove that for a domain $A$, $A^n$ cannot have (an isomorphic copy of) $A^m$ as a subring where $m>n$. My attempt is as follows: If we let $\{b_i\}_{i=1}^n$ be a basis for $A^n$ and ...
0
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0answers
24 views

Exact sequence associated to a acyclic complex

I am reading chapter V of 'Homological Algebra' by Cartan and Eilenberg. Regarding a module $A$ as a complex with $A^0=A$, $A^n=0$ for $n\neq 0$ and zero differentiation. The augmentation ...
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1answer
34 views

How to compute the homomorphism module?

I want to compute the homomorphism module $\textrm{Hom}_{\mathbb{Z}}(\mathbb{Z} /{p^{n}}, \mathbb{Z} /{p^{m}})$ for $m\leq n$. Can someone please help me!
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1answer
31 views

An odd expression appearing in proof that kernel and image of certain map on group ring $\mathbb Z[G]$ are equal

Let $G = \langle g \rangle$ be a cyclic group of order $n$. Consider the free ring $\mathbb Z[G]$ of all formal sums of elements from $G$ with coefficients from $\mathbb Z$, with multiplication given ...
3
votes
1answer
43 views

Why are Grothendieck's and Hartshorne's definitions of quasi-coherence equivalent?

Hartshorne's Algebraic Geometry defines an $\mathcal O_X$-module $\mathscr F$ to be quasi-coherent if there is an open affine cover $(U_i=\operatorname{Spec} A_i)_{i\in\mathcal I}$ of $X$ such that ...
1
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2answers
47 views

Example of a module which is free at an isolated point

I'm looking for the most simple example of a quasicoherent sheaf $\mathcal{F}$ over a scheme $X$ (preferably affine for simplicity) which has a free stalk $\mathcal{F}_x$ at a point $x \in X$ and yet ...
2
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2answers
50 views

“Unique simple object” of category of modules, simplicity and irreducible modules

I can't seem to find a decent definition of "unique irreducible module" for an algebra. In what sense is the uniqueness? For example, does $\mathfrak{sl}_2$ have a unique irreducible module, since it ...
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1answer
36 views

Some questions about standard $K$-duality

Let $A$ be a finite dimensional $K$-algebra, where $K$ is an algebraically closed field. We define a funtor $D: mod A \to mod A^{op}$ called standard $K$-duality. Suppose that $M$ is an arbitrary ...
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1answer
40 views

In what sense are complete local rings finitely generated modules?

In the first paragraph of section 18.4 of Eisenbud's Commutative Algebra, there is the following comment. Most interesting Noetherian rings can be written as finitely generated modules over ...
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0answers
32 views

projective resolution of finitely generated modules

I am in the condition where I have a noetherian ring $R$ of finite global dimension. Consider the category of finitely generated (right) modules over $R$. Then I want to show that every module admits ...
1
vote
1answer
15 views

ACC on chains of finitely generated submodules

Noetherian rings are those having ascending chain condition on ideals. There are also literatures concerning ACC on n-generated ideals (i.e., generated by n elements); see e.g, Commutative Rings with ...
5
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1answer
76 views

Tensor product of $\mathbb{Q}$ with an infinite product [closed]

How can I prove that the tensor $\mathbb{Q} \otimes \left( \prod_n \mathbb{Z}/n\mathbb{Z} \right)$, where the product is taken over all the positive integers $n$, is not trivial?
3
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1answer
38 views

Verify proof that if $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian

I have to prove that If $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian We let $S$ be a non-finitely generated submodule of $M\otimes_R ...
0
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0answers
17 views

Torsion-free submodule of maximal rank

Let $A$ be a free $\mathbb{Q}[t_1^{\pm 1},\dots,t_\mu^{\pm 1}]$-module of rank $n$. If $M$ is a submodule of $A$, I know that it must be torsion free. Assuming the rank of $M$ is $n$ can I conclude ...
0
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0answers
31 views

General definition of free module over arbitrary rings and space of multi-linear functions

Consider definition of free module as given in this question: Free modules over commutative ring (possibly without unity) where free means having a LI spanning set I have proved if Can we say if V ...
2
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1answer
78 views

Free modules over commutative ring (possibly without unity) where free means having a LI spanning set

Let us define free module over a ring (possibly without unity) as: Def: M is said to be free module over ring R (possibly without unity) if there exist X subset of M such that X is LI and spans M. ...
2
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1answer
18 views

Z modules spanned by row space of matrix invariant under matrix multiplication

I have met this strange looking problem on which I have no idea, from my course on Abstract Algebra dealing with modules: Let $ v_1,...,v_k \in \mathbb{Z}^n $ row vectors of length n over $ ...
2
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0answers
31 views

Extending ring metric to arbitrary module

I am working on my own on an idea I had and this one is a bit difficult so I am seeking a bit of inspiration and to see what already exists. Let $R$ be a given ring with a metric $d$ and $M$ being an ...
2
votes
3answers
70 views

Is $A / \mathfrak{m}$ flat if $A$ is a local ring?

I'd like to prove the following: if $A$ is a local ring and $\mathfrak{m} \subset A$ its maximal ideal, then $A / \mathfrak{m}$ is a flat $A$-module. How can I do this? I've tried to find a suitable ...
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2answers
78 views

Submodules and quotients of finitely generated modules

Let $G$ be a finitely generated abelian group whose free part has rank $r$. I know that every subgroup $H$ is finitely generated and has free part of rank $s\leq r$, and that also $G/H$ is finitely ...
3
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1answer
21 views

Is an injective map of $B$-modules also injective as an $A$-linear map if $B$ is an $A$-algebra?

I've been going through my submitted exercises again of my Commutative Algebra-class and I have the following question: Let $A$ be a commutative ring with unity. Given any injective homomorphism of ...
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0answers
27 views

Constructing a ring consisting of formal infinite series from a given ring

Let $A$ be an $\mathbb{N}$-graded $\Bbbk$-algebra, where $\Bbbk$ is a field, and where $\dim_\Bbbk A_n < \infty$ for all $n \in \mathbb{N}$. I can't see anything preventing me from constructing a ...
0
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2answers
47 views

Prove that $\operatorname{Ext}^{d+1}(A, B)\cong \operatorname{Ext}^1(M_d,B)$

So, given a resolution, with $P_{i}$ projective modules: $$0\longrightarrow M_d\longrightarrow P_{d-1} \longrightarrow \cdots \longrightarrow P_0 \longrightarrow A\longrightarrow 0,$$ I'm trying to ...
7
votes
2answers
131 views

Problem with the ring $R=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q &\Bbb Q\end{bmatrix}$ and its ideal $D=\begin{bmatrix}0&0\\ \Bbb Q & \Bbb Q\end{bmatrix}$

Let us consider the ring $ R:=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q & \Bbb Q\end{bmatrix} $ and its two-sided ideal $ D:=\begin{bmatrix}0 & 0\\ \Bbb Q & \Bbb Q\end{bmatrix} $. Let then ...
0
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2answers
24 views

Quotient space decomposition

Let $V$ be a vector space(or module) with decomposition $V= V_1 \oplus V_2$. And let $W \subset V$ be a subspace with decomposition $W= W_1 \oplus W_2$ such that $W_1 \subset V_1$ and $W_2 \subset ...
0
votes
2answers
37 views

Checking commutativity of a diagram of modules over some ring and what the commutativity of the diagram implies.

Suppose that you have the following diagram of modules over some ring: These are my questions: (1) To prove that the diagram is commutative, we needs to prove that $gf=kh$, $wf=rv$, $zh=uv$, ...
3
votes
1answer
49 views

Category of modules

For given a ring $R$, we can define the category of left $R$-modules. In fact, objects are all left $R$-modules and morphisms are $R$-module homomorphisms. NOw my question is: If we do not fix $R$, ...
2
votes
1answer
32 views

Kernel of a natural map is a direct summand of the covariant extension

I am reading chapter 2 of 'Homological Algebra' by Cartan and Eilenberg. 1/ Given a ring homomorphism $\varphi: \Lambda \rightarrow \Gamma$ and a right $\Gamma$-module $A$, we can treat $A$ and ...
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0answers
41 views

Free $R$-module when $R$ is not unital

We can easily construct free $R$-module when $R$ is unital by setting $$R[S] = \{ f\colon S\to R\,|\, f\ \text{finitely supported}\}$$ and defining operations pointwise. The key here is that we can ...
1
vote
2answers
65 views

Definition of exact sequence [closed]

Take a sequence $A \to B \to C \to 0$ of modules over a commutative ring. How would one show that it is exact? I understand the necessary surjectivity of $B \to C$, but what about the first map?