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

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1answer
25 views

Ideal as projective module

Let $R$ be a commutative ring without zero divisors. Assume that ideal $a\subset R$ is a projective $R$-module. How to prove that $a$ is finitely generated ? I need only hints.
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1answer
18 views

Exact sequences of projective modules

Let $0{\rightarrow} K \stackrel{g}\rightarrow P\stackrel{f}\rightarrow Q \rightarrow 0$ and $0\rightarrow K' \stackrel{g'}\rightarrow P'\stackrel{f'}\rightarrow Q \rightarrow 0$ be exact sequences ...
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1answer
29 views

Ideals and injective modules

Let $I$ be a left ideal of $R$. Assume that there exist element in $I$, which is not a zero divisor. How to prove that for every (left) injective $R$-module $Q$ we have $IQ=Q$ ? I need only hints. ...
2
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1answer
48 views

When is $M\otimes N$ a module?

So suppose we have a $R$-$S$-bimodule $M$, and an $S$-$T$ bimodule $N$, then we can construct the abelian group $M\otimes_S N$. Under what conditions could we make this a module? I would expect this ...
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2answers
25 views

Injective modules

Let $\{R_n\}_{n\in \mathbb{N}}$ be a collection of fields. Let $R:=\prod\limits_{n=1}^{\infty}R_n$. Show that for every $n$ $R_n$ is an injective $R$-module, but $\bigoplus\limits_{n=1}^{\infty}R_n$ ...
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1answer
32 views

Prove that there's a unique morphism that completes the commutative diagram

I have to prove that there's a unique $\gamma : M'' \rightarrow N''$ that completes this diagram considering the rows are exact. $$\begin{array} MM' \stackrel{f_1}{\longrightarrow} & M & ...
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1answer
18 views

Another approach to $A/I\otimes_A A/J\simeq A/(I+J)$?

The same question appears here $A/ I \otimes_A A/J \cong A/(I+J)$ however I'm looking for a different approach. Let $A$ be an algebra, $I$ a left ideal and $J$ a right ideal. I'd like to show ...
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1answer
15 views

Can one extent to (left) modules on IBN ring the results of vector spaces?

We know that (left) modules "are not as nice as vector spaces" because lots of properties of these ones don't fit to the most general modules. But il we restrict to modules on IBN ring (i.e. modules ...
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1answer
15 views

Ring of polynomials as free module

Is it true that $R=k[x,y]$ is a free $R$-module ? I think that it isn't true. Natural candidate for the base is $\{x^{\alpha}y^{\beta}\}_{\alpha,\beta}$, but : $x\cdot (xy) +(-y)\cdot x^2 =0$ and ...
3
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0answers
18 views

Computing simplicial homology via Smith Normal Form over Rings

I am not sure whether this is the right forum to ask such a question, if not please let me know. In the context of my masters thesis, I am working on writing a program to compute simplicial homology ...
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1answer
27 views

Find all the possible pairs of submodules $M_1$ and $M_2$ of the $\mathbb{Z}$-module $\mathbb{Z}_{18}$ so that $\mathbb{Z}_{18} = M_1 \oplus M_2$

I started by considering the possible proper subgroups of $\mathbb{Z}_{18}$, which are $\langle\bar{9}\rangle$, $\langle\bar6\rangle$, $\langle\bar3\rangle$ and $\langle\bar2\rangle$, which are also ...
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1answer
72 views

$\operatorname{Hom}_R(\mathfrak{a},M)$ is isomorphic to $\mathfrak{a}^{-1}M$ if $R$ is a Dedekind domain

I want to prove Lemma 2.5.1 of Silverman's Advanced Topics in The Arithmetic of Elliptic Curves (whose proof is left to the reader): Let $R$ be a Dedekind domain, let $\mathfrak{a}$ be a ...
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1answer
13 views

Radical of the annihilator of an element of a Noetherian module

Assume $M$ is a commutative Noetherian $R$-module and $m\in M$ is such that $P=\sqrt{\operatorname{Ann}(m)}$ is a prime ideal in $R$. Is it true that $P$ is an associated prime of $M$, i.e. there is ...
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2answers
36 views

Characterizing the A-module M/S

I've been working through this for a little while, and I'm not 100% sure I understand what I'm supposed to be doing here, or maybe I'm not grasping correctly what they mean by "Characterize". ...
0
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1answer
12 views

About extensions of finitely generated modules.

Let $R$ be a ring with unity, and $A \subseteq B$ be $R$-modules. I want to know if the following is true: If $A, B/A$ are finitely generated $R$-modules, then $B$ is finitely generated. Under ...
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3answers
155 views

A simple example of a ring that is an $A$-module but not an $A$-algebra

Let $A$ and $B$ be commutative rings, and suppose (the underlying group of) $B$ has a structure of $A$-module. "Obviously", that doesn't imply that $B$ gets a structure as an $A$-algebra, but I can't ...
4
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1answer
68 views

Under what conditions does $M \oplus A \cong M \oplus B$ imply $A \cong B$?

This question is fairly general (I'm actually interested in a more specific setting, which I'll mention later), and I've found similar questions/answers on here but they don't seem to answer the ...
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1answer
43 views

What is the difference between submodules of $A/\mathfrak a$ as an $A$-module or as an $A/\mathfrak a$-module?

If $\mathfrak a$ is an ideal of unital commutative ring $A$, Then we can see to $A/\mathfrak a$ as an $A$ module or as an $A/\mathfrak a$ module. If $A=\mathbb Z$ there is no structure difference ...
4
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0answers
61 views

Example of $A$-module but not $A$-algebra. [duplicate]

If $A$, $B$ are commutative rings, and if $B$ is an $A$-algebra then it is also an $A$-module. I am looking for an example that shows that the converse is not true. That is, I am looking for ...
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3answers
98 views

All simple modules are projective $\Rightarrow$ semisimple [duplicate]

Let $A$ be a finite dimensional algebra over a field $K$. It is clear that if $A$ is semisimple, then every simple module is projective. Does the converse hold ? It seems false, but I can't find a ...
4
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1answer
121 views
+50

What is $\operatorname{Hom}_R(P,R)$ isomorphic to when $P$ is projective?

Let $R$ be a (possibly noncommutative) ring with $1$. Now, quite clearly we have $$\operatorname{Hom}_R(R^n,R)\cong R^n.$$ I am wondering if there is any similar result for ...
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1answer
37 views

$M_1$, $M_2$ and $M_1\cap M_2$ injective imply $M_1+M_2$ is also injective?

Let $M$ be an $R$-module ($R$ is a ring with identity) and let $M_1$ and $M_2$ be two injective submodules such that $M_1\cap M_2$ is also injective. How to show $M_1+M_2$ is injective? If the ...
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0answers
21 views

Fiber product of $f$ and $g$ is isomorphic to $\mathbb Z\oplus \mathbb Z_p$?

Let $p$ be a prime number. I'm supposed to show the fiber product (pullback) of the canonical projections $f:\mathbb Z\longrightarrow \mathbb Z_{p}$ and $g:\mathbb Z_{p^2}\longrightarrow \mathbb Z_p$ ...
5
votes
2answers
311 views

How to prove surjectivity part of Short Five Lemma for short exact sequences.

Suppose we have a homomorphism $\alpha, \beta, \gamma$ of short exact sequences: $$ \begin{matrix} 0 & \to & A & \xrightarrow{\psi} & B & \xrightarrow{\phi} & C & \to & ...
0
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1answer
13 views

module isomorphism inbetween two equivalence classes of polynomials

Let $g \in \mathbb{R}[t]$ be a normed irreducible polynomial of degree 2, meaning that $g(t) = (t - \lambda)(t - \overline{\lambda}$) for a $\lambda = a + b i$, with $a, b \in \mathbb{R}$, $b ≠ 0$. I ...
1
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1answer
34 views

How to define a group ring when the group is infinte?

I am trying to prove the following from Tadao Oda's "Convex bodies and algebraic geometry" Let $N \cong \mathbb Z^n$ be a free $\mathbb Z$ - module of rank $n$ and $M = \text{Hom}_{\mathbb ...
3
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1answer
72 views

global dimension of rings and projective (flat) dimension of modules

Let $R$ be ring such that every left $R$-module has finite projective dimension ( resp. finite injective dimension). Is the left global dimension of $R$ finite? Similarly, Let $R$ be ring such that ...
3
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1answer
45 views

Difference between $(N+P)/N$ and $P/N$

If $N$ and $P$ are submodules of the $A$-module $M$ (where $A$ is a commutative ring with unity), why is there a difference between $(N+P)/N$ and $P/N$? If $x\in (N+P)/N$ then $x=n+p+N=p+N$ for some ...
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2answers
36 views

Prove that $M$ is finitely generated and it is a semi-simple module.

Let $M$ be a $R$-module. It is given that intersection of all maximal sub-modules of $M$ is the zero module. Moreover the module is given to be Artinian. Prove that $M$ is finitely generated ...
2
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3answers
37 views

Prove that if $M$ is finitely generated then it is Artinian.

Let $M $ be a semisimple $R$-module. Prove that if $M$ is finitely generated then it is Artinian. To show this we have to prove that every non-empty collection of sub-modules of $M$ has a minimal ...
4
votes
1answer
58 views

Two modules are isomorphic in the stable module category iff they are projectively equivalent

Let $R$ be a (not necessarily commutative) ring. Let ${\text{mod-}R}$ be the category of finitely generated right $R$-modules. Let $\underline{\text{mod-}R}$ be the stable module category, with the ...
2
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1answer
33 views

why is a simple ring not semisimple?

A simple module is a semisimple module . A module $M$ is called semisimple if every submodule is a direct summand of $M$ Since a simple module has $\{0\}$ and $M$ as its submodules so it is ...
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0answers
26 views

Graphs associated with rings and modules

There are several articles in the literature that deals with some interesting graphs associated with rings and modules. For example The zero-divisor graphs D. F. Anderson, P. S. Livingston, The ...
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2answers
28 views

Showing that a specific surjective $R$-module homomorphism is an isomorphism [closed]

Suppose that $M$ is a Noetherian $R$-module and $f$ is a surjective $R$-module homomorphism from $M$ to $M$. Is $f$ an isomorphism?
6
votes
1answer
126 views

Prove $AB=I$ implies $BA=I$ using Fitting's Lemma

I know this question has already been asked, but I need a proof that for $A,B \in M_n(K)$, $$AB=I_n \Rightarrow BA=I_n$$ using Fitting's lemma. I thought of using the fact that $K^n$ is a ...
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2answers
20 views

Do modules have to be defined over rings with unity?

This is definition for left module over a ring $R$ given in Wikipedia: Suppose that $R$ is a ring and $1_R$ is its multiplicative identity. A left $R$-module $M$ consists of an abelian group ...
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1answer
24 views

A question from the proof of: If M is a free f.g R-module, and N is a submodule of M, then N is free, where R is a PID

I'm reading the proof from Dummit and Foote third edition, page 460-461,theorem 4. I am confused about one thing: On page 461, there are: (a) $M=Ry_1\oplus \ker v$ (b)$N=Ra_1y_1\oplus (\ker v\cap ...
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0answers
8 views

Projective space of a module

I'm studying projective geometry in a basic course of geometry. My question is: Is there an equivalent definition of projective space not of a vector space but of a module? I think the basic ...
3
votes
2answers
49 views

relation of R-module homomorphisms with direct sums

If $f$ is a module homomorphism from $A$ to $B$ and $g$ is a module homomorphism from $B$ to $A$, and $g \circ f$ is the identity function, why is it that $B= \operatorname{im}(f) + \ker(g)$?
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1answer
427 views

Trivial extension of an algebra

Suppose that $A$ is a finite dimensional $k$-algebra. Call $Q=\mathrm{Hom}_k(A,k)$. $Q$ admits an $A$-$A$-bimodule structure in the obvious way. The trivial extension of $A$ is defined as follows: ...
2
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2answers
60 views

Characterization of projective modules?

I'm having a hard time with a characterization of projective modules: A $R$-module $P$ is projective if and only if for every epimorphism $f:I\longrightarrow I^{\prime\prime}$ with $I$ injective and ...
2
votes
1answer
23 views

Finitely generated and free module

We work in the category $R-\mathsf{Mod}$ where $R$ is unital. Some authors define free modules to have a finite basis. If we don't require a basis to be finite, I think it is quite obvious that a ...
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1answer
33 views

Every projective $R$-module $P$ is free

I have come across a theorem which states that if the underlying ring $R$ is a principal ideal domain then every $R$-module $P$ which is projective is free also. But the problem is I have encountered ...
17
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4answers
636 views

Pathologies in module theory

Linear algebra is a very well-behaved part of mathematics. Soon after you have mastered the basics you got a good feeling for what kind of statements should be true -- even if you are not familiar ...
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0answers
39 views

How bad must be a ring to allow cyclic artinian modules that are not noetherian?

I've been studying the relations between artinian and noetherian modules over commutative rings. One can prove two interesting results for the commutative case. Theorem Every commutative artinian ...
0
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1answer
25 views

Inverting a nonzerodivisor of a module

I'm reading the Paper "What makes a complex exact?" by Eisenbud and Buchsbaum. On page 266 it says: Thus we may assume $0 \neq \operatorname{rank}(\phi_n,L) < \operatorname{rank}(F_n)$ and ...
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0answers
33 views

Left Ideals and Two-Sided Ideals of Rings of Matrices

This is a question I posted in this topic: Pathologies in "rng". However, I decided that the question deserves its own thread, and I want to know if anybody can answer it. For ...
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3answers
34 views

Is there an easy example of a module which is not projective

Is there an easy example of a module which is not projective? I found that direct product of $\mathbb Z$ is not projective but its proof is complicated. Are there any easy examples for this?
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0answers
50 views

What are the ring theoretic properties of this trivial extension? [closed]

Let $R={\mathbb Z }_{ (p) }{ \ltimes\mathbb Z }_{ { p }^{ \infty } }$ be the trivial extension of ${\mathbb Z }_{ (p) }=\{ a/b\in\mathbb Q:b\neq 0,\gcd(a,b)=1\text{ and } p\nmid b\}$ by ${\mathbb Z ...
2
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0answers
21 views

How to show that direct product of $\mathbb Z\times\mathbb Z\times\mathbb Z\times…$ is not projective as a $\mathbb Z$ module? [duplicate]

How to show that direct product of $\mathbb Z\times\mathbb Z\times\mathbb Z\times...$ is not projective as a $\mathbb Z$ module? I know that $\mathbb Z$ is a free $\mathbb Z$ module since it has ...