Questions tagged [projective-module]

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

Filter by
Sorted by
Tagged with
4
votes
2answers
94 views

Are projective modules over $\mathbb{Z}[x_1,…,x_m]$ free?

The Quillen-Suslin theorem states that any finitely generated projective module over $\mathbb{k}[x_1,...,x_m]$ is free, for $\mathbb{k}$ a field. Is it known whether this statement is true in the case ...
0
votes
1answer
38 views

The duality on projective modules takes minimal presentations to minimal presentations

I've been working through Auslander, Reiten, and Smalø's Representation Theory of Artin Algebras. In reading through the basic setup relevant to the transpose operation, I've come across two ...
0
votes
1answer
21 views

R is projective Q-module?

The set of real numbers $\mathbb{R}$ is project $\mathbb{Q}$-module? I think it is not but I cannot prove it. How can I prove it?
2
votes
0answers
41 views

$K$-group of category of bounded chain complexes of free modules with finite length homologies

For a Noetherian local ring $(R, \mathfrak m)$, let $K_0^{\mathfrak m}(R)$ denote the abelian Group generated by isomorphism classes of bounded chain complexes of finitely generated free modules with ...
0
votes
1answer
46 views

Proof of Quillen's patching theorem

The following are from Lam's book Serre's problem on projective modules, on page 163 and 164. For any ring $A$, the notation $m \in \mathscr{R}^A(A[t_1, \dots, t_n])$ means that there exists a $A-$ ...
0
votes
1answer
20 views

If $f:R\to S$ is an $R$-algebra and $P$ is a projective $S$-module, then $pd_R(P)\le pd_R(S)$.

The title basically says it all. If $f:R\to S$ is an $R$-algebra and $P$ is a projective $S$-module, then I need to show that $pd_R(P)\le pd_R(S)$. Here $pd_R(A)$ refers to the projective dimension of ...
1
vote
1answer
59 views

Simple tensors and projective modules [closed]

For a (unital) ring $R$, let $M$ and $N$ be a projective right, and left, $R$-module respectively. Does it hold that $0 \neq m \otimes n \in M \otimes_R N$ for all non-zero $m \in M$, $n\in N$? If so, ...
0
votes
0answers
49 views

Topological K-theory and characteristic classes of module bundles

Let $R$ be a commutative ring and let $A$ be a commutative unital topological $R$-algebra. By means of replacing vector spaces with $A$-modules, one can define $A$-module bundle in analogy to vector ...
2
votes
0answers
27 views

Second summand to make projective module free

Suppose there's a projective $R$-module $P$ (non-free). We know that there is another $R$-module $M$ such that $P\oplus M$ is free over $R$. Is there a way to write down such an $M$ in terms of $P$? ...
1
vote
0answers
29 views

Projective cover of simple counts occurences of simple in composition series?

On page 11 in Tensor Categories is the statement Let $\mathcal{C}$ be a finite abelian $k$-linear category. Then for any $X,Y \in \mathcal{C}$ with $X$ simple we have \begin{align*} \dim_k \...
1
vote
1answer
35 views

Tensoring with projective module preserves injectivity.

Let $A$ be a ring, $P$ a projective left $A$-module and $E,F$ two right $A$-modules. If $u:E\rightarrow F$ is an injective homomorphism, the homomorphism $$u\otimes 1_P:E\otimes_A P\rightarrow F\...
2
votes
1answer
23 views

Finitely generated projective resolution

Let $K$ be a field, $A$ be a finite dimensional $K$-algebra and $M$ be a finitely generated $A$-module. Is it true that $M$ admits a projective resolution by finitely generated projective $A$-modules?
1
vote
0answers
41 views

Does an Equivalence of Finitely Generated projective module categories over ring of functions imply a diffeomorphism between smooth manifolds?

Let $M$ be a smooth connected manifold with $R = C^{\infty}(M,\mathbb{R})$ the $\mathbb{R}$-algebra of smooth real-valued functions on $M$. Let $ProjFinMod_R$ denote the category of finitely generated ...
0
votes
1answer
17 views

Flat Module and free torsion

I need prove the follow assertion: "Let A be a integer domain ring. If M is a flat A-module then M is free-torsion" My definition of flat module is: a R-module F is flat if the functor $ F \...
1
vote
0answers
17 views

Total number of nonisomorphic uniserial modules for a Nakayama algebra

Let $\Lambda$ be a Nakayama algebra and $\Lambda=\coprod\limits_{i=1}^sn_iP_i$ with $P_i$ nonisomorphic indecomposable projective modules ($n_iP_i$ denotes $n_i$ copies of $P_i$). Prove that the total ...
0
votes
0answers
34 views

Why does the injective module have no projective summand?

Sorry if this is too elementary. The problem is from Auslander's Representation Theory of Artin Algebras, page 214 proposition 5.6. Let $\Sigma$ be an artin algebra(gl.dim$\Sigma$=dom.dim$_{\Sigma}\...
0
votes
1answer
27 views

The module $\text{Hom}_C(E,F)$ of two finitely generated projective $C$-modules

Let $C$ be an abelian ring and $E,F$ two finitely generated projective modules. Then $\text{Hom}_C(E,F)$ is a finitely generated projective $C$-module. First of all, since $C$ is abelian, the ...
0
votes
0answers
23 views

If $gcd=(m,n)=1$ I have to prove that the ring $\mathbb{Z}_m$ is projective over the ring $\mathbb{Z}_{mn}$

I have to prove that the ring $\mathbb{Z}_m$ is projective over the ring $\mathbb{Z}_{mn}$ and also I need to prove that $\mathbb{Z}_m$ is not a free module over $\mathbb{Z}_{mn}$ . I only have the ...
1
vote
1answer
33 views

Homomorphic image of a projective module

I just encountered one statement out of blue when reading section 20 of 'local representation theory ' by Alperin. Every module is a homomorphic image of a projective module. Not sure how to see ...
1
vote
1answer
25 views

Can nonisomorphic indecomposable projective modules contain isomorphic projective submodules?

Let $P_1$ and $P_2$ be nonisomorphic indecomposable projective $R-$modules. Can they contain isomorphic projective submodules? The answer would be extremely helpful for my studies of basic algebras.
2
votes
2answers
47 views

Equivalent properties of integral domain f.g projective modules

Question : Let $R$ be an integral domain and $K$ be its fraction field. Let $M$ be an $R$ module. Then the following are equivalent: (1) $M$ is projective $R$-module such that $[M\otimes_R K : K]$ ...
1
vote
0answers
28 views

Show that the modules are nonisomorphic

Defenition: Let $\Lambda$ be an algebra. $\Lambda$ is called basic if $\Lambda \simeq \coprod\limits_{i=1}^nP_i$, where $P_i$ are nonisomorphic indecomposable projective modules. Problem: Let $\...
1
vote
1answer
34 views

Are these the projective and injective representations of these quivers?

Find all indecomposable projective and indecomposable injective representations of these two quivers over the field $k$ up to isomorphism. I've drawn my answer in this picture. Can you please check ...
1
vote
1answer
36 views

When does a finitely generated projective module surjects onto the ring

Let $M$ be a non-zero, finitely generated projective module over a commutative integral domain $R$. Is it necessarily true that there is an $R$-linear surjection $M\to R$ ? If this is not true in ...
2
votes
0answers
37 views

Are finite projective modules locally free without axiom of choice?

Let $A$ be a commutative ring. An $A$-module $M$ is called a finite projective module if there exists a module $N$ and a natural number $n\in\mathbb N$ such that $M\oplus N\cong A^n$, and an $A$-...
0
votes
1answer
83 views

$\rho:P \to M$ a projective cover for $M$ semi-perfect module. Prove $M$ is indecomposable if and only if $P$ is indecomposable.

Let $M \neq \lbrace 0 \rbrace$ be a semi-perfect module and $\rho:P \to M$ a projective cover for $M$. I want to prove that $M$ is indecomposable if and only if $P$ is indecomposable. All right, we ...
1
vote
1answer
47 views

Extension of a projective submodule along a surjective map

Suppose that a surjection of commutative noetherian rings $A \to B$ is given. Suppose you have a free $B$-submodule of $B^n$ for some $n$. You can easily extend this to a $A$-submodule of $A^n$ by ...
3
votes
1answer
42 views

Projective modules and lifting property

I am trying to understand projective modules while avoiding as much as possible the use of exact sequences and category theory. The definition I am using is that a module $P$ is projective if it is ...
1
vote
0answers
19 views

Vector bundle with space of sections Finite projective module

Let $B$ be a $T_{3\frac12}$ space, $C(B)$ be the ring of continuous real valued functions on $B$. Show that if $M$ is a finitely generated projective module over $C(B)$, then there is a vector bundle $...
0
votes
1answer
36 views

Prove that if $X/A$ is a free module and if $A$ is a submodule of $X$, then $A$ is a direct summand of $X$ [closed]

Prove that if $X/A$ is a free module and if $A$ is a submodule of $X$, then $A$ is a direct summand of $X$. I need to prove it without sequences.
1
vote
1answer
34 views

The property $m \otimes_A n = 0$ for modules over an algebra [closed]

For $A$ a not necessarily commutative algebra over a field $\mathbb{k}$, can there exist two bi-modules $M$ and $N$ over $A$ such that, for some $m \in M$, and $n \in N$, $$ m \otimes_A n = 0? $$ I ...
1
vote
0answers
42 views

Isomorphism of Projective Modules

Good folks! Having problem sorting out the solution to a question that should be relatively simple from Weibel's book of $K$-theory. Hoping you could help me. The question in particular is Exercise 2....
0
votes
0answers
64 views

Existence of Projective Cover

Let $A$ be a finite dimensional algebra and let $M$ be a finite dimensional $A$-module. Then $M$ has a projective cover and I am trying to prove this. Let $\mathcal{C}$ be a module class and $X$ a ...
0
votes
1answer
32 views

Infinite direct sum of projective modules

Let $A$ be a noncommutative algebra over a field $k$, and $M$ a left $A$-module such that $$ M = \bigoplus_{i \in I} M_i $$ where $I$ is a countable set, and the decomposition is a decomposition of ...
0
votes
1answer
23 views

Constructing a resolution from other projective resolutions and a short exact sequence

Suppose we have a short exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ and a projective resolution $P$ for $A$ and a projective resolution for $C$, Q. Is there a way we can ...
1
vote
1answer
45 views

Constructing the projection of a projective module from its dual basis

Let $M$ be a (left) module over a not necessarily commutative algebra $A$. We can say that $M$ is projective if, for some $n \in \mathbb{N}$, i) there exists a projection $P$ (i.e. $P \in M_n(A)$ ...
2
votes
1answer
51 views

Kernel of a morphism between projective modules and base change

Let $A$ be a ring (Noetherian if it helps) and $f : M \to N$ be a morphism between finite projective $A$-modules. Let $B$ be an $A$-algebra. Is it true that if $f\otimes_A B$ is injective then $\ker(...
0
votes
1answer
34 views

About the Properties of Perspective Projection to Lower Dimension

I know this is loose and generic, but I just need to check my logic and if there is any reference or method ye may refer to, will be very much appreciated. I have a function $f(\vec{v})$ which only ...
1
vote
1answer
34 views

$\operatorname{pd}(M)\leq n\implies\operatorname{Ext}_{R}^{n+1}(M,N)=0$

Let $M,N$ be $R$-modules. Since $\operatorname{pd}(M)\leq n$, we have a projective resolution, $$\cdots\rightarrow0\rightarrow P_n\rightarrow\cdots\rightarrow P_0\rightarrow M\rightarrow0.$$ Then $\...
0
votes
0answers
16 views

If (co)domain of map of chain complexes is in(pro)jective, then every quasi-isomorphism is a homotopy equivalence

I'm reading some lecture notes which are an introduction to homotopy theory, and there is a short section on chain complexes, where the difference between a quasi-isomorphism and homotopy equivalence ...
0
votes
1answer
36 views

Projective Modules, Definition and Etymology

Question that I have after reading some of the answers to this question about the etymology of projective modules. It goes as follows: Let $P$ be an $R$-module, $R$ being some arbitrary ring. Is it ...
0
votes
0answers
23 views

Finding the rank of the endomorphism algebra of a projective module over a non-commutative algebra

Let $A$ be a commutative ring, $B$ be a non-commutative $A$-algebra, $P$ be a finitely generated projective left $B$-module. Let $E=\text{End}_B(P)$ as a left $A$-module (by restricting from $B$). If ...
2
votes
1answer
69 views

Two definitions for regular rings

A noetherian ring $R$ is said to be regular if every localization at a prime ideal is regular local. On the other hand, there is another definition of regularity for non-noetherian rings: A (...
4
votes
2answers
71 views

How do we prove intuitionistically that free modules of finite rank are projective?

I'm struggling to understand how the following proof can be intuitionistically valid$^1$: Theorem. Let $R$ be a commutative ring. A free module of finite rank $R^n$ is projective. Proof. Let $...
2
votes
3answers
57 views

$M=\mathbb{Z}/2018\mathbb{Z}$ is not a projective $\mathbb{Z}$-module

$M=\mathbb{Z}/2018\mathbb{Z}$ is not a projective $\mathbb{Z}$-module. I know that $\mathbb{Z}$ is a PID and therefore any projective $\mathbb{Z}$-module is also free. $M$ is not a free $\mathbb{Z}$-...
4
votes
1answer
96 views

Projective resolution of $k$ as $k[x]/(x^n)$-module?

I'm trying to get some familiarity with projective resolutions and therefore working on this little exercise: Find a projective resolution of a field $k$ considered as a $k[x]/(x^n)$-module (where $x$...
1
vote
1answer
52 views

$R$ is a PID and $M=R/(d)$ an $R$-module. Show that $M$ is not projective.

Note: $(d)$ is a nonzero proper ideal of $R$. My attempt: Consider $\pi:R\to R/(d)$, then $\ker(\pi)=(d)$. I want to prove that there is no $\sigma:R/(d)\to R$ such that $\pi\circ \sigma=id_M$. ...
2
votes
1answer
39 views

$\operatorname{Hom}_A(A^*, P)\cong P$ such that $P$ is projective under the given assumption?

Let $A$ be a finite dimensional algebra over a field $k$. Let $A^* = \operatorname{Hom}_k (A,k)$ and $A^*$ is $A$-bimodule. Suppose that $\operatorname{Hom}_A (A^*, A) \cong A$ as left $A$-module. ...
0
votes
1answer
43 views

Find the factor groups of the $\mathbb{Z}/252\mathbb{Z}$ which are projective $\mathbb{Z}/252\mathbb{Z}$-modules

I'm reading about projective modules and saw an algebra exam problem that and uses a bunch of concepts I am familiar with (question in the title) but I am not sure of the best method to solve it. ...
3
votes
1answer
75 views

Projective ideal in $\mathbb{Z}[\sqrt{-5}]$

This is a question from my past Qual "Set $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$. Is $I$ prime? Is $I$ projective as an $R$-module?" Clearly $R/I \cong \mathbb{Z}/2\mathbb{Z}$, hence $I$ ...

1
2 3 4 5
12