Questions tagged [projective-module]

For questions related to projective modules, their structures, and properties.

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$As + At = A$ $\implies$ $as^k+bt^k=1$ [duplicate]

So I am studying this proof of Quillen's theorem. (page no. 27) Here they say for a commutative ring $A$ we are given $As+At=1$. So we get $as+bt=1$ for some $a,b\in A$. But, in the proof they used ...
Nilotpal Chakraborty's user avatar
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Determinant of one stably free projective module

A projective $A$-module $P$ is called one stably free if $P\oplus A\simeq A^n$ for some $n\in \mathbb{Z}$. It can be seen that such projective modules are in one-to-one correspondence with the orbits ...
Varadharajan R's user avatar
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Projective covers

Let $R$ be a ring with $1$ and $M$ be a right $R$-module. A homomorphism $f:P\to M$ is said to be a projective cover of $M$ if $P$ is projective and $f$ is an epimorphism and $\ker f \ll P$. Recall ...
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Projective and free modules

In the graded context, if $R = K[x_1,...,x_n]$ where $K$ is a field, is a projective R-module a free R-module?
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Free summand of a module vs. finite direct sum of copies of it

Let $M$ be a finitely generated module over a commutative Noetherian ring $R$. If $R$ is a direct summand of $M^{\oplus n}$ for some $n\ge 1$ , then is $R$ a direct summand of $M$? I can easily prove ...
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Projective ideals of $\frac{\mathbb{Z}}{180\mathbb{Z}}$

Determine all the projective ideals of the ring $\frac{\mathbb{Z}}{180\mathbb{Z}}$. Give an example of an ideal that is not projective. We can begin by applying the CRT to find some of the projective ...
Tim's user avatar
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Finitely presented modules and homomorphism modules

Suppose $M$ and $N$ are two finitely presented $R$-modules, where $R$ is a commutative ring. Suppose in addition that $M$ is a projective $R$-module. How does one show that $\mathrm{Hom}_R(M,N)$ is a ...
BillyJohny's user avatar
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Prove that $\mathbb{C}$ is not a projective $\mathbb{Z}$-module

Prove that $\mathbb{C}$ is not a projective $\mathbb{Z}$-module. Suppose on the contrary that $\mathbb{C}$ is a projective $\mathbb{Z}$-module. Then we have that $\mathbb{C}$ is a submodule of $\...
Tim's user avatar
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Are module finite algebras over Noetherian semiperfect rings again semiperfect?

Let $S$ be a Noetherian semiperfect ring (https://en.m.wikipedia.org/wiki/Perfect_ring). Let $R$ be a module finite associative $S$-algebra. Then, is $R$ also a semiperfect ring? (Clearly, $R$ is ...
uno's user avatar
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If $a^{3}=a$ for some non-zero element $a$ in $R$, then the ideal $Ra$ is projective $R$-module

I need help to the following problem: Problem: Let $R$ be a commutative ring with $1 \neq 0$. Let $a$ be a non-zero element of $R$ such that $a^3=a$, then the ideal $Ra$ is a projective $R$-module. ...
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Trouble understanding this paragraph on projective modules

I have been learning about modules in algebra for a few days now using Nicholson's Abstract Algebra, (I am a slow reader, it has taken me hours over the course of several days just to go through 7 ...
iwjueph94rgytbhr's user avatar
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Let A be a submodule of a projective R-module B, show that A is projective if B/A is projective

I want help to the following problem(especially I want to tell me which is the basic idea behind the solution): Problem: We have the following assumptions: Let $B$ is projective $R-$module $A$ is $R-...
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$\mathrm{Hom}_A(P(1),S(1))$, $\mathrm{Hom}_A(S(1),S(1))$ and $\mathrm{Hom}_A(S(1),I(1))$ as vector spaces

Let $A$ be a finite dimensional $K$-algebra where $K$ is an algebraic ally closed field. Since $A$ finite dimensional, the right $A-$module $A_A=e_1A\oplus e_2A\oplus\cdots \oplus e_nA$ is an ...
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If $M$ is a finitely generated module over a local ring, then $M$ is free.

Given a local ring $A$ and a finitely generated projective module $M$, show that $M$ is also free. So I'm actually required to prove this question, although I believe that I managed to prove that $M$ ...
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How Should I Prove that M is Torsion-free Given that M is Projective?

Question: R is an integral domain, M is a projective R-module and Tor(M) = {$x\in M$: rx = 0 for some non-zero $r\in R$}. Prove that M is torsion-free (i.e. Tor(M) = {$0_M$}). Attempt: I don't seem to ...
Mr Prof's user avatar
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Projective module over a commutative ring $R$ becomes free over some ring containing $R$ .

I am reposting this as earlier question was probably not clear to some. So I was reading a book in which we have a commutative ring $A$, a commutative algebra over it called $B$ and a set of ...
Tony Pizza's user avatar
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Why a nonzero finite abelian group is not projective?

Here is the question I am trying to solve (I know it is answered here $A$ be a nonzero finite abelian group then $A$ is not a projective or injective $\Bbb Z$ module. but the answer is not very clear ...
Intuition's user avatar
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Incorrect proof for $e$ idempotent $\Rightarrow$ $eA$ projective

I have seen several proofs for $e$ idempotent $\Rightarrow$ $eA$ projective where $A$ is an algebra. I tried something different and produced a proof without using the fact that $e$ is idempotent (so ...
kubo's user avatar
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$M_1 \oplus M_2$ is projective $\iff M_1$ and $M_2$ are both projective

If $M_1$ and $M_2$ are R-modules, $R$ possibly non-commutative. Show that $M_1 \oplus M_2$ is projective $\iff M_1$ and $M_2$ are both projective I have no idea how to solve this. By definition an $R$-...
darkside's user avatar
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Tensor product of projective modules.

How do we prove tensor product of two projective modules is projective? My attempt: Let $P ,P'$ be projective modules. By equivalent conditions we have $K ,K'$ such that $K \oplus P$ and $K' \oplus P'...
Ezed's user avatar
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Are injective maps of finite modules over regular ring liftable to termwise injective maps of finite projective resolution?

Let $R$ be a regular noetherian ring and let $M$ be a finitely generated $R$-module. In this situation, we know that $M$ admits a finite resolution in terms of finite projective $R$-modules. Now, let $...
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Is every projective R-module torsion-free? when R is a noncommutative ring! [duplicate]

Let R be a noncommutative ring, is every projective R-module torsion-free? or, for what kinds of rings (or algebras) the answer is affirmative?
Kami sh's user avatar
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partial Quillen-Suslin on square matrix

The Quillen-Suslin theorem states that projective modules over $\mathbb{Q}[x_1, \dots , x_n]$ are free. (This also holds over more general fields or rings.) I have a square $m\times m$ matrix $M$ over ...
MatthysJ's user avatar
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Are projective modules free over a polynomial ring with infinite indeterminates over a field?

(1) In 1963, Bass proved that any nonfinitely generated projective module is free over a connected Noether ring. (2) Quillen–Suslin theorem states that any finitely generated projective module over a ...
Liang Chen's user avatar
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How Can We Prove Flatness from an Induced Exact Sequence?

I want to prove that for every short exact sequence $$ O \to A \xrightarrow{f} B \xrightarrow{g} C \to O $$ of $R$-module homomorphisms, if the induced sequence $$ O \to M\otimes_R A \xrightarrow{\...
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Is Every Isomorphism on Tensor Product of Modules an R-module Isomorphism?

We have an R-module M. We also have S, a multiplicatively closed subset of R. I want to prove that there exists a unique R-module isomorphism f: S$^{-1}$R$⊗$M $\to$S$^{-1}$M, defined by f(${\frac rs}$⊗...
Mr Prof's user avatar
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1 answer
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If a Short Exact Sequence is Split-Exact, Does that Mean it is Left-Split?

If a short exact sequence is split-exact, can we conclude it is left-split? Motivation: I am asking this question because I want to prove that if every $R$-module is projective, then every $R$-module ...
Mr Prof's user avatar
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How Should I Prove this Function is Well-defined?

I am trying to prove that the dual sum of P$_1$ and P$_2$ is projective iff P$_1$ and P$_2$ are projective. I am done with every aspect of the proof. I have this diagram which commutes: Now, let h: P$...
Mr Prof's user avatar
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Is this a valid proof of Well-definedness?

We are given $L, M$ and $N$ as unitary $R$-modules, and $f:M\rightarrow N$ as an isomorphism. We need to prove that $f^*:Hom_R(N,L)\rightarrow Hom_R(M,L)$ is an isomorphism. I started by defining the ...
Mr Prof's user avatar
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How Should I Prove this Function is an Isomorphism?

I have this question: Let $L$, $M$ and $N$ be unitary $R$-modules. Let $f: M \to N$ be an $R$-module isomorphism. Prove that the map $f_*: \text{Hom}_R(N,L) \to \text{Hom}_R(M,L)$ is an isomorphism. ...
Mr Prof's user avatar
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Standard projective resolution of module of the path algebra

I have the quiver $Q$ with two vertices $v_1,v_2$ and two arrows $a_1,a_2$ that go from $v_1$ to $v_2$. Let $kQ$ be the path algebra of my quiver, where $k$ is a field, and let $M=kQ/\langle a_1\...
kubo's user avatar
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Double dual basis for projective modules

It is standard that an $A$-module $P$ is projective iff. there exist elements $p_i \in P$ and $\bar{p}_i \in \hom_A(P, A)$ for $i$ in some indexing set $I$ such that for all $q \in P$, we have $$q = \...
Bubaya's user avatar
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Show that the statements about module $M_R$ are equivalent.

I am trying to prove the equivalence of the following statements regarding Module $M_R$. I have no idea about the mechanism of proof in such statements: Let $A, L\in M_R, \alpha:A \rightarrow L, B \...
Emad kareem's user avatar
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1 answer
42 views

Is this Proof of Well-definedness Correct?

I am trying to prove that the function $\bar g$ is well-defined. This is how I did it: Define $\bar g: \ker\alpha \to \ker\beta$ by $\bar g(m) = g(m)$, for all $m$ in $\ker\alpha$. This means $m$ is ...
Mr Prof's user avatar
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1 answer
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Schanuel's Lemma

I want to prove the following: Assume that we have two exact sequences of $R$-modules $$0 \to K \xrightarrow{\psi} P \xrightarrow{\varphi} M \to 0$$ and $$0 \to K' \xrightarrow{\psi'} P' \xrightarrow{...
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$P$ is projective implies $P℘$ is a free $R℘$-module

Statement: Let $R$ be a commutative Noetherian ring and let $P$ be a finitely generated $R$-module, then $P$ is projective if and only if $P℘$ is a free $R℘$-module for all $℘$ in $Spec(R)$. For the ...
Nilotpal Chakraborty's user avatar
1 vote
1 answer
60 views

Isomorphism of projective modules modulo nilpotent ideals.

Let $R$ be a commutative ring, $I$ be an ideal of $R$ with $I^2 = 0$ and let $M, N$ be projective $R$-modules. Then $M/IM$ and $N/IN$ are projective $R/IR$-modules and we get a map $\mathrm{Hom}_R(M, ...
Candyblock's user avatar
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Show that if $Q$ is a non-trivial $\mathbb{Z}$-module, that is divisible, then $Q$ can not be a projective $\mathbb{Z}$-module. [duplicate]

As the question says, I want to show that If $Q$ is a non-trivial $\mathbb{Z}$-module, that is divisible, then $Q$ can not be a projective $\mathbb{Z}$-module. Now, here is my reasoning: Let $F$ be ...
Ben123's user avatar
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A finite abelian group $A$ can not be a projective $\mathbb{Z}$-module.

Any $\mathbb{Z}$-module $P$ is projective if it is a direct summand in some free $\mathbb{Z}$-module $M$. Now, if $M$ is some free $\mathbb{Z}$-module, then $M \cong \bigoplus_{a \in A} \mathbb{Z}$ ...
Ben123's user avatar
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Trouble understanding torsion

I'm reading this paper. In definition 6.1.2 it is mentioned that $H_n(G;A) \cong \text{Tor}_n^{\mathbb{Z}G}(\mathbb{Z}, A)$ (for all $n=0, 1, 2, ...$). What exactly is meant by the right-hand side? I'...
JMM's user avatar
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Projective resolution of $\mathbb{Z}/p^n\mathbb{Z}$ on $\mathbb{Z}/p^k\mathbb{Z}$

Consider $R=\mathbb{Z}/p\mathbb{Z}$ for $p$ prime and $n>1$ as ring and $M_k=\mathbb{Z}/p^k\mathbb{Z}$ as $R$- module ($k \leq n$). I want to find a projective resolution of $M_k$. $R$ is ...
Mario's user avatar
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Projective module on the ring of dual numbers

Let $\mathbb{K}$ be a field and consider the ring of dual numbers $R=\mathbb{K}[x]/<x^2>$. I have to prove that any projective $R$-module $P$ is injective. My idea is to use the Baer’s criterion....
Mario's user avatar
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4 votes
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74 views

The stable category of $\mathbb{Z}$

Is there an alternative description/characterization of the stable module category of Abelian groups? I guess that the category of torsion groups is a subcategory of it, but is it all of it? What is ...
Michal's user avatar
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stable range of stably free modules

This is part of exercise 1.1.5 of the K-book: Notation: we say $R$ has stable range at most $n$ if every unimodular row $(r_0,\ldots, r_n)$ induces a unimodular row $(r_1',\ldots, r_n')$ with $r_i'=...
SummerAtlas's user avatar
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3 votes
1 answer
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Noetherian, self-injective ring $R$ with non-torsionless $R$-module

Let $R$ be an (associative, unital) ring. By an $R$-module we mean a left $R$-module. We call an $R$-module torsionless if it can be embedded into a direct product of the regular $R$-module $R$. I am ...
Margaret's user avatar
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Projective module base change

Let $M$ be a finitely generated projective $C[0,1]$-module. Without using any vector bundle theory, how does one show that the pullback of $M$ along the evaluation at $0$ and evaluation at $1$ maps ...
Anupam's user avatar
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What is the importance of a projective cover and injective hull for a module?

My understanding of projective covers and injective hulls for modules over a (finite-dimensional) associative $\mathbb{C}$-algebra $A$ is as follows. $\bullet$ The projective cover of $M$ is an ...
Cornelius's user avatar
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Confused about projective modules and locally free modules

I'm trying to read an arXiv paper using the terms projective and locally free modules interchangeably within the context of finitely generated number fields. To even attempt to understand the results ...
Inspector gadget's user avatar
1 vote
1 answer
57 views

Does a power of endomorphism factor through a projective?

Let $A$ be a ring (e.g. an $R$-algebra over a noetherian local ring $R$), $M$ an indecomposable $A$-module, and $f$ an endomorphism, which is not an isomorphism. Now, we assume that the endomorphism ...
sola's user avatar
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an isomorphism of dual module

$R$ is a commutative ring with identity. $M^*={\rm Hom}_R(M,R)$ is the dual module of $M$. It is known that there is an $R$-homomorphism $M^*\otimes_R N\to {\rm Hom}_R(M,N)$. My question is the ...
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