The modules tag has no wiki summary.

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**1**answer

431 views

### Is $\mathbb{R}$ a $\mathbb{C}$-module without AC?

Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial(scaler multiplication is not identicaly zero) $\mathbb{C}$-module.
Now my questions are?
0.Is it consistent with $ZF$ that $\mathbb{R}$ is ...

**3**

votes

**1**answer

124 views

### Canonical presentation of pro-modules over pro-rings

Let $A = (\dotsc \twoheadrightarrow A_2 \twoheadrightarrow A_1 \twoheadrightarrow A_0)$ be a (commutative) pro-ring with surjective transition maps. Consider the category $\mathcal{M} := \varprojlim_i ...

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**0**answers

89 views

### Ring of endomorphisms as a criterion of a dimension 1 module

Let $R$ be a unital ring and $M$ be an $R$-module. I have some questions about relation between the ring $\operatorname{End}_R M$ of endomorphisms and the notion of “dimension” of a module. I’m not ...

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**0**answers

366 views

### Submodule embeddable in a finitely generated module

I have a terminology question. For a commutative ring $A$ (not necessarily Noetherian), $A$-modules that are isomorphic to an $A$-submodule of a finitely generated $A$-module form a fairly good class ...

**4**

votes

**1**answer

163 views

### $ mult(R/I) = d_1 \cdots d_r \quad \Rightarrow \quad f_1,\dots,f_r \quad \text{is a $R$-regular sequence?}$

We define multiplicity of a module M of dimension $d>0$ as $$mult(M) := lc (P_M) (d-1)!$$ where $P_M$ denotes the Hilbert polynomial of M. Equivalently, we have $mult(M) = Q_M(1)$, where $HP_M (z) ...

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votes

**1**answer

167 views

### Is a projective module of constant finite rank finitely generated?

If $R$ is a (commutative) ring and $P$ is a projective $R$-module, then every localization of $P$ at a prime of $R$ is free by Kaplansky's theorem, and has a well-defined rank. If these ranks are all ...

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**0**answers

44 views

### Is there a positive integer k such that any endomorphism of any free module over any commutative ring is a linear combination of k idempotents?

Consider the following Condition (C) on a positive integer $k$:
(C) If $R$ is a commutative ring, if $F$ is a free $R$-module, and if $f$ is an endomorphism of $F$, then $f$ is an $R$-linear ...

**-3**

votes

**1**answer

86 views

### Isomorphic quotient of a Module over Noetherian commutative algebra [closed]

I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve ...

**1**

vote

**1**answer

117 views

### General criterion to find a Z-basis in a fixed generating subset

Let $V=\mathbf{Z}^N$ be a free $\mathbf{Z}$-module of rank $N$. Let $S\subseteq V$
be a fixed finite subset.
Consider the submodule $M:=\langle S\rangle\leq V$ generated by $S$. We know form the ...

**2**

votes

**0**answers

49 views

### an equality in module theory

Let $R$ be a ring and $M$ be an $R$-module. Let $N$ and $L$ be two submodules of $M$ such that
$(rad(N+L):M)=\sqrt{(N:M)+(L:M)}$. Then can we conclude that $(N+L:M)=(N:M)+(L:M)$?
Note that ...

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**0**answers

196 views

### R[[X]] flat as a R[X]-module?

I assume $R[X]\rightarrow R[[X]]$ is not flat in general, but I was wondering if any conditions on a commutative ring $R$ are known such that $R[[X]]$ is flat as a $R[X]$-module.
Would $R$ noetherian ...

**6**

votes

**0**answers

88 views

### Finiteness for separated residually finite modules

Suppose that $A$ is a commutative noetherian Jacobson ring and $M$ is an $A$-module. Suppose in addition that $M$ is $\mathfrak{m}$-adically separated for every maximal ideal $\mathfrak{m}$, and that ...

**5**

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**1**answer

172 views

### Injective flat module

Let $R$ be a (right noetherian) ring. Is there always a right $R$-module which is both flat and injective? If $R$ is an integral domain, then the answer is indeed yes, as the quotient field is such.
...

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vote

**2**answers

253 views

### A semisimple group ring

Let $n \in \mathbb{N}$, $p$ a prime number, and $G$ a finite group of order coprime to $p$. Let $R = \mathbb{Z} /p^n \mathbb{Z}$ be the ring of integers mod $p^n$. Must $R[G]$ be semisimple?
As noted ...

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**0**answers

354 views

### An exact sequence which does not split

Let $X$ and $Y$ be indecomposable modules over a finite dimensional algebra and let $f \colon X \to Y$ be a non-zero morphism which is neither a monomorphism nor an epimorphism.
Suppose that it is ...

**2**

votes

**1**answer

57 views

### dg-flat complexes and their characters

Let $\otimes$ denotes the usual tensor products of complexes and symbols live in the category of chain complexes of $R$-modules. Let $X$ be a dg-flat complex (i.e. $X_n$ is flat for each n and ...

**-3**

votes

**1**answer

222 views

### Doubt in this proof of Horrocks theorem

I'm beginning to study some research papers and I need right now to understand the solution of Vaseršteĭn of Serre's theorem (simplest proof of this theorem), to do so, I'm beginning to understand ...

**11**

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**7**answers

697 views

### Properties of rings that have an elegant description in terms of the associated category of modules

Suppose $A$ is a ring. Then $A$ happens to be a division ring iff every left $A$-module is free. (See here for proofs). I think this is very beautiful; what other properties of rings have an elegant ...

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**1**answer

148 views

### Complexity of rational $\mathrm{GL}_{n(r)}$-modules

Let $k$ be an algebraically closed field of characteristic $p>0$, and let $G=\mathrm{GL}_n(k)$ for some natural number $n$. For any integer $r\ge 1$, let $G_{(r)}$ denote the $r$th Frobenius ...

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**0**answers

97 views

### Intuitive idea: What is the motivation for relative homological algebra

Basically I was wondering what is the purpous of choosing a smaller class of epis in a category? Does it allow for the existence of "more" "projective-like" objects?
What are some applications? For ...

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votes

**1**answer

272 views

### Re-interpreting vector spaces in a choice-less model of ZF as modules over a regular ring in ZFC

I am searching a module $M$ over a (von Neumann) regular ring $A$ ($\forall a\in A$, $\exists x\in A$: $axa=a$) with two properties:
(1) every finitely generated submodule of $M$ is projective ...

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votes

**1**answer

628 views

### Why are they called Specht Modules?

I know that the simple modules of $\mathbb{C}S_n$ are called Specht Modules, and they are named after the German Mathematician Wilhelm Specht because he studied them, but I think these modules were ...

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**1**answer

144 views

### Simultaneous decomposition of modules over Dedekind domains

I posted the question on mathexchange as well, but realized that my chances would be higher posting here;
In the paper "Almost diagonal matrices over Dedekind Domains" by L. Levy a specific ...

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votes

**1**answer

71 views

### Injective modules over noncommutative noetherian rings

Let $R$ be a left noetherian ring, then it is well known that Matlis proved that any injective $R$-module decomposes into a sum of indecomposable injectives, each of form $E(R/I)$ for some irreducible ...

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**1**answer

262 views

### Splitting as $\mathbb{F}_p[[X]]$-modules

Let $A$ be a finitely generated torsion $\mathbb{Z}_p[[X]]$-module, $B$ = { $x \in A$ such that $px=0$ } and $C=A/B$ where $\mathbb{Z}_p$ denotes the $p$-adic integers. Given $ 0 \rightarrow B/pB ...

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votes

**0**answers

92 views

### Sum of projective submodules of a projective over a semihereditary ring

Sorry in advance if this is too silly. Let $R$ be a right semihereditary ring and $P$ a projective right $R$-module. It is well-known that finitely generated (thus projective) submodules of $P$ form a ...

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votes

**2**answers

281 views

### Dual of a module

Let $M$ be a $ \mathbb{Z}_{p}[[T]] $-module and $X=Hom(M,\mathbb{Q}_{p}/\mathbb{Z}_{p})$ be the dual of $M$. Let $X[p^n]$ denotes the $p^n$-torsion points of $X$. Is $X/X[p^n]$ the dual of $M[p^n]$ ...

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**2**answers

407 views

### Does base extension reflect the property of being isomorphic?

Let L/K be a (separable?) field extension, let A be a finite dimensional algebra over K, and let M and N be two A-modules. Let $A' = L \otimes_K A$ be the algebra given by extension of scalars, and ...

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**1**answer

233 views

### Is this square commutative?

Suppose that the following commutative diagram of $R$-modules is given.
$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ ...

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**1**answer

265 views

### Maximal Submodule of a Verma Module

Let $\mathfrak{h}$ be a Cartan subalgebra of a $\mathbb{C}$-semi simple Lie algebra $\mathfrak{g}$. Given $\lambda \in \mathfrak{h}^*$, $M(\lambda)$ the Verma module of highest weight $\lambda$ and ...

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votes

**5**answers

744 views

### is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles?

This might be obvious to experts, but I'm not sure where to look for the answer. On a reasonably nice, at least noetherian, scheme (or variety, algebraic space, stack), can the category of coherent ...

**1**

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**1**answer

137 views

### arrows in the injective representations of quivers

Let $Q$ be a quiver (a directed graph) and $E$ be a representation of $Q$ by R-modules, that is, for each vertex $v$ in $Q$ there exists an R-module $E(v)$ and for each arrow $a:v\to w$ there exists a ...

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**1**answer

373 views

### Splitness of commutative diagrams

Consider the following commutative diagram in the category of $R$-modules where $R$ is an associative ring with identity and all modules are unital.
$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ ...

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128 views

### Is this a pure monomorphism?

Let $M$ be a representation of a quiver $Q=(V, E)$ by $R$-modules. By $M^{+}$ we mean a representation of $Q^{op}$ with $M^{+}(v)=\mathrm{Hom}(M(v), \frac{Q}{Z})$. One can easily see that there is ...

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**1**answer

207 views

### finitely presented representations

Let Q=(V,E) be a direct graph where V is the set of all its vertices and E denotes the set of all its arrows. $X$ is called a representation of Q by modules if it is a functor from Q to R-Mod. i.e. ...

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vote

**1**answer

207 views

### Pure monomorphism of functors-

Suppose that $F, G: Q\rightarrow R{\rm -Mod}$ be two covariant functors where $Q$ is an abelain category and $R$ is a commuatative ring. Also let $\eta: F\rightarrow G$ be a natural transformation ...

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**1**answer

242 views

### Analogy between Lagrange's Theorem and Rank-Nullity Theorem?

One can view view Lagrange's Theorem $$|G/H|=|G|/|H|$$ and the Rank-Nullity Theorem $$\dim(V/U)=\dim(V)-\dim(U)$$ as directly analogous. Does anyone know a high-level explanation of this analogy? I ...

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**2**answers

115 views

### On modules with finite uniform dimension

Is it true that if a module $M$ has finite uniform dimension then the same is true for its homomorphic images ?

**8**

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**1**answer

205 views

### Rings in which every module has an injective image

Consider the class of rings $R$ with identity such that any left $R$-module has a non-zero injective homomorphic image. Any such ring is clearly a left V-ring. Is it true that any such ring must be ...

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**2**answers

216 views

### Looking for a reference: double orthogonal complement in $(\mathbb{Z}/q\mathbb{Z})^n$

I'm using the following result in a computer science paper:
Let $V$ be a submodule of $(\mathbb{Z}/q\mathbb{Z})^n$ (n-tuples with addition and multiplication mod $q$). Let
$$V^\perp = \{u \in ...

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**3**answers

441 views

### Is it true that simple projective modules are injective?

It is known that simple modules are either projective or singular. Is it true that simple projective modules over (commutative) rings are injective ?

**8**

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**3**answers

450 views

### What is the categorical significance of the trivial $\mathfrak{g}$-module in the category of $\mathfrak{g}$-mod?

This question may be trivial for experts.
Let $\mathfrak{g}$ be a Lie algebra over a field $k$ and consider the category $\mathfrak{g}$-mod of $\mathfrak{g}$ modules. We can add suitable conditions, ...

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**0**answers

227 views

### Epimorphisms between external tensor products

Let $k$ be a field, $R,S$ commutative $k$-algebras. If $\mathsf{Mod}_{fp}$ denotes the category of finitely presented modules, the external tensor product $\mathsf{Mod}_{fp}(R) \otimes_k ...

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**1**answer

99 views

### pure sub-complexes of exact subcomplexes

In https://www.google.com/#q=tensor+product+of+complexes%2Benochs a new tensor product of complexes is defined which characterizes flatness in the category of complexes of $R$-modules. That is, a ...

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**1**answer

177 views

### cofree modules and dual

1, Why do people pay special attention to Q/Z in the definition of cofree modules instead of ordinary abelian groups?
2, Over a PID, is every injective module cofree? Just like the relationship ...

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votes

**1**answer

134 views

### is every finitely n-presented (S^{-1})R-module a localization of a finitely n-presented R-module?

Let S be a multiplicative set in a ring R. We can see that every finitely generated $(S^{-1})R$-module is a localization of a finitely generated R-module.
Then, more generally, is every finitely ...

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vote

**1**answer

169 views

### Classification of pairs of commuting endomorphisms

Let $K$ be an algebraically closed field. I'm interested in isomorphism classes of triples $(V,f,g)$ where $V$ is a finite dimensional $K$-vector space and $f,g$ are commuting endomorphisms of $V$. ...

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130 views

### simple tensor product of modules over algebras

Let $M$, $N$ be simple modules over associative algebras $A$ and $B$ (over $\mathbb{C}$), respectively. When is $M\otimes N$ simple as a $A\otimes B$-module?
It is right if $A$ or $B$ has a ...

**-1**

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**1**answer

164 views

### Pure Quotient and pure sub-object

Let $\mathcal{C}$ be the category of modules over a ring.
Let also $\mathcal{F}$ be a class of objects in $\mathcal C$ closed under pure subobject (pure quotient) and direct limit. Is $\mathcal{F}$ ...

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225 views

### Rejects and injectives

Let $A$ be any ring and consider modules on the left. For $M$ $A$-module, the trace $Tr(M,A)$ is a two-sided ideal of $A$. If $A$ is a unitary ring then:
$Tr(P,A)P=P$, for $P$ projective;
...