Skip to main content

Questions tagged [projective-module]

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

Filter by
Sorted by
Tagged with
0 votes
1 answer
39 views

${\mathrm{Ext}}_{k[\epsilon]}^1(k, k)$ and ${\mathrm{Ext}}_{k[X]}^1(k, k)$.

For a field $k$, I am calculating ${\mathrm{Ext}}_{k[\epsilon]}^1(k, k)$ and ${\mathrm{Ext}}_{k[X]}^1(k, k)$, where $\epsilon^2 = 0$. However, there seems no complete explanation as far as I checked. ...
Pierre MATSUMI's user avatar
1 vote
0 answers
43 views

Module over local ring which is free after cutting down by a non-zero-divisor

Let $(R,\mathfrak m)$ be a commutative Noetherian local ring and $M$ be an $R$-module such that $M\neq \mathfrak m M$. Let $x\in \mathfrak m$ be a non-zero-divisor on both $R$ and $M$. If $M/xM$ is a ...
Alex's user avatar
  • 433
0 votes
0 answers
26 views

Tensor product of two modules that are projective over different rings

This is in regards to the proof of Weibels Corollary 9.1.5: Suppose $A$ is a projective module over a comm. ring $k$. I have already proven, that this implies $A^{\otimes n}$ is also $k$-projective. ...
Vacarn's user avatar
  • 1
1 vote
0 answers
35 views

cokernel of acyclic chain complex of projective modules after tensoring

Let $R$ be a commutative Noetherian ring, $(\mathbf F,d)$ be an acyclic chain complex of projective $R$-modules and $M$ be a finitely generated $R$-module. Let $i$ be an integer and put $N:=\text{...
Alex's user avatar
  • 433
-4 votes
1 answer
64 views

Establish a short exact sequence $0\to Z_p\to Z_{p^2}\to Z_p\to 0$ does not split, every submodule of the projective module need not be projective. [closed]

Establish a short exact sequence $0\to Z_p\to Z_{p^2}\to Z_p\to 0$ does not split, every submodule of the projective module need not be projective.
Haval Mohammed's user avatar
0 votes
1 answer
53 views

Equivalent conditions of $\textrm{Ext}^2(A,B) = 0$ for all $A$ and $B$

I am studying homological algebra and I am having difficulty proving the equivalences of the following: (i) If $0 \to A \xrightarrow{f} B$ is exact and $B$ is projective, then $A$ is projective (ii) ...
Squirrel-Power's user avatar
0 votes
0 answers
54 views

If any direct product of the regular module is a submodule of a projective module, then is any direct product of the regular module projective?

Let $R$ be a ring with unity. As is well known, $R$ is left perfect and right coherent if and only if any product of projective left modules is projective. If $M$ is a left torsionless module over $R$,...
Liang Chen's user avatar
0 votes
1 answer
28 views

If $M$ is a finitely generated module and $\operatorname{pd}(M) = n < \infty$, then is there a finitely generated projective resolution of $M$?

Let $R$ be a ring with unity, and let $M$ be a finitely generated left module over $R$ with projective dimension $\operatorname{pd}(M)=n<\infty$. Is there a projective resolution of $M$ of the form ...
Liang Chen's user avatar
2 votes
1 answer
38 views

If a ring $R$ decomposes as $_R R \simeq P_1 \oplus \cdots \oplus P_n$, can any simple $R$-module be isomorphic to a quotient of some $P_i$?

If $R$ is a left Artin ring with unity and with an arbitrary decomposition of the regular module $_R R \simeq P_1 \oplus \cdots \oplus P_n$, then we can conclude that any simple module over $R$ must ...
Liang Chen's user avatar
1 vote
0 answers
18 views

If $R$ is a ring with unity, can the regular module $_R R$ always be decomposed into a direct sum of indecomposable projective modules? [duplicate]

If $R$ is a left Artin ring with unity, we can conclude that the regular module $_R R$ can be decomposed into a direct sum of indecomposable projective modules. Since $R$ is left Artin, $_R R$ is a ...
Liang Chen's user avatar
1 vote
0 answers
54 views

Cancellation property of vector bundles

Let $R$ be a Noetherian ring of dim d and $ Y=Spec R[X_1,\ldots, X_n]= SpecR \times_{SpecZ} Spec Z[X_1,\ldots, X_n]$. Consider $U$ to be locally free sheaf of finite type over Y with rank $\geq max\{d+...
Dgarg12's user avatar
  • 91
1 vote
0 answers
19 views

Chasing diagrams of chain complexes which are pointwise projective

For context this question arose in a simple K-theory computation, where I wish to show that the inclusion of the category of bounded degreewise projective chain complexes into the category of bounded ...
DevVorb's user avatar
  • 1,495
1 vote
0 answers
27 views

If $X \oplus Y$ has a projective cover denoted $P$, can we find projective covers for $X$ and $Y$ respectively? [duplicate]

Let $X, Y$ be two modules over a ring with unity $R$. If $X$ and $Y$ have projective covers denoted $P_X$ and $P_Y$ respectively, then $X \oplus Y$ has a projective cover $P_X \oplus P_Y$. This is ...
Liang Chen's user avatar
2 votes
1 answer
36 views

If $P$ is an indecomposable projective module over a ring with unity $R$, can $P/rP$ also be an indecomposable module over $R$?

Let $P$ be a non-zero projective module over a ring with unity $R$, and let $r$ be the Jacobson radical of $R$. If $P/rP$ is indecomposable, then $P$ is indecomposable. This is because if $P = P_1 \...
Liang Chen's user avatar
1 vote
1 answer
52 views

Projective Kernel or Image of Homomorphism between Projective Modules

Note: A simple "yes" or "no" answer would be sufficient. I just need confirmation. Let $R$ denote a Noetherian commutative ring (e.g. $\mathbb{C}[x]/(x^2)$). If $A,B$ denote ...
JAG131's user avatar
  • 917
2 votes
1 answer
56 views

Does there exist a projective module $P$ such that $P/rP$ is a simple module, yet $\pi: P \to P/rP$ fails to act as a projective cover map of $P/rP$?

If $P$ is a projective module over a semiperfect ring $R$ such that $P/rP$ is a simple module over $R$, then the canonical surjection $\pi: P \to P/rP$ must be a projective cover map of $P/rP$, where $...
Liang Chen's user avatar
1 vote
1 answer
48 views

A Question Concerning the Properties of a Projective Cover

If $R$ is a left perfect ring, then any module $M$ over $R$ possesses a projective cover denoted by $P$. Furthermore, we can deduce that $P/\text{Rad} P$ is a semisimple module, where $\text{Rad} P$ ...
Liang Chen's user avatar
3 votes
1 answer
86 views

If $N$ contains a maximal submodule, does this imply that $ \text{Rad} N $ is a superfluous submodule of $N$?

If $M$ is a finitely generated module over a ring $R$ with unity, it can be readily demonstrated that the radical of $M$, denoted $ \text{Rad} M$, qualifies as a superfluous submodule of $M$. The ...
Liang Chen's user avatar
2 votes
0 answers
50 views

Extensions of $G$-modules parametrized by $H^1$

Let $G$ be a finitely generated group and let $V$, $W$ be one-dimensional representations of $G$ over $\mathbb{F}_q$. (I guess we can think of $V$ and $W$ simply as $G$-modules, which are isomorphic ...
Conjecture's user avatar
  • 3,270
0 votes
1 answer
35 views

Clarification about stable module category

I'm trying to read about stable module categories, and in the definition they say, "$f \sim g$ if $f − g$ factors through a projective module", where $f$ and $g$ are module homomorphism. I'm ...
StuckInTheFridge's user avatar
0 votes
1 answer
45 views

If $Ext(P,\_) = 0,$ then $P$ is projective

Trying to understand Bredon's proof (below). Please note that $G$ is an Abelian group. So my confusion starts with $H_1(Hom(R,G_*)) \to H_0(Hom(A,G_*)) \to H_0(Hom(F,G_*))$. I understand that he is ...
strugglingmathguy's user avatar
1 vote
1 answer
72 views

If $P$ is projective, then $Ext(P,\_)=0$

My attempt: $P$ is projective. Take $0 \to B \to I_0 \to I_1 \to 0$ is an injective resolution of $B$. Apply $Hom(P,\_)$, left exactness of covariant functor implies exactness of the sequence: $0 \to ...
strugglingmathguy's user avatar
1 vote
1 answer
36 views

Rigidity of finitely generated projective modules

I am currently reading a lecture notes on module theory and more specifically on Morita Theory. Here is a porition of the lecture note that I do not understand. Finitely generated projective modules ...
Squirrel-Power's user avatar
4 votes
0 answers
117 views

Given a projective $R$-module $M$, how does one find another $R$-module $N$ such that $M \oplus N$ is a free $R$-module?

Given a projective $R$-module $M$, how does one find another $R$-module $N$ such that $M \oplus N$ is a free $R$-module? In my case, $R = {\mathbb{C}[x, y]}/{(y^2 - x^3 + x)}$, and $M$ is the ideal ...
Noam's user avatar
  • 51
1 vote
0 answers
39 views

Generation of a projective module by fiber product

Proposition: Let $R$ be a commutative noetherian ring and let $P$ be a finitely generated $R$-module. Then $P$ is projective if and only if $P_m$ is free for all maximal ideals $m$ of $R$. Statement: ...
Nilotpal Chakraborty's user avatar
2 votes
1 answer
68 views

Is fiber product of $M_s$ , $M_t$ over $M_{st}$ isomorphic to $M$?

Let M be an R-module, where R is a commutative ring with unity. Take two elements $s,t \in R$ such that $Rs+Rt=R$. Then we want to show below commutative diagram is fiber product diagram: Basically ...
Nilotpal Chakraborty's user avatar
0 votes
1 answer
66 views

$As + At = A$ $\implies$ $as^k+bt^k=1$ [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
2 votes
1 answer
108 views

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
1 vote
0 answers
75 views

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 ...
Hussein Eid's user avatar
  • 1,071
0 votes
0 answers
109 views

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?
Cib's user avatar
  • 37
1 vote
0 answers
66 views

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 ...
uno's user avatar
  • 1,560
0 votes
0 answers
48 views

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
  • 303
1 vote
1 answer
96 views

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
2 votes
1 answer
82 views

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
  • 303
2 votes
0 answers
70 views

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
  • 1,560
0 votes
1 answer
53 views

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. ...
TrItOs's user avatar
  • 111
0 votes
1 answer
77 views

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
1 vote
1 answer
65 views

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-...
TrItOs's user avatar
  • 111
0 votes
0 answers
38 views

$\mathrm{Hom}_A(P(1),S(1))$, $\mathrm{Hom}_A(S(1),S(1))$ 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 ...
fusheng's user avatar
  • 1,159
1 vote
1 answer
142 views

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 $M$ ...
Ubik's user avatar
  • 488
0 votes
1 answer
162 views

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
  • 451
2 votes
0 answers
67 views

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
0 votes
2 answers
136 views

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
  • 3,127
0 votes
1 answer
48 views

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
  • 2,067
2 votes
3 answers
141 views

$M_1 \oplus M_2$ 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
  • 589
1 vote
1 answer
152 views

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
  • 51
0 votes
0 answers
20 views

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 $...
Stabilo's user avatar
  • 1,538
0 votes
0 answers
22 views

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
  • 53
2 votes
0 answers
34 views

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
  • 111
2 votes
0 answers
381 views

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

1
2 3 4 5
17