1
vote
1answer
42 views

A condition of equivalence of flatness and projectiveness

This is a problem in "Foundations of Module and Ring Theory" of Wisbauer: " Let $R$ be a subring of the ring $S$ containing the unit of $S$. Show that a flat $R$-module $N$ is projective if and only ...
4
votes
2answers
119 views

Proving that tensoring a projective module with a flat module gives a projective module?

If $P$ is a projective module and $M$ is a flat module, both over some commutative ring $R$, then how do you prove that $M\otimes_R P$ is flat? I tried using the fact that $P$ is the direct ...
5
votes
0answers
49 views

When is an $R$-projective module a projective module?

Let $R$ be a semiperfect ring. Is it a true fact that every $R$-projective module $M$ with $Rad(M)$ superfluous in $M$ is projective? I could not reach a good result using just the fact that ...
1
vote
1answer
23 views

Morita equivalence: Is $_{\mathrm{End}_R(P)}P$ projective if $P_R$ is?

Assume $P$ is a right projective $R$-module. Is $P$, viewed as a left $\mathrm{End}_R(P)$-module, projective as well? If not, under what conditions does it hold? Context: I am trying to ...
2
votes
1answer
43 views

Starting projective modules problems

Show that if $n=rs$ where $n,r,s>1$ are positive integers, then the $ \mathbb{Z}_n $-module $r \mathbb{Z}_n$ is projective but it is not free if $(r,s)=1$. Any ideas or help how to prove this ...
0
votes
1answer
32 views

Every finitely generated module is the quotient of a finitely generated projective module.

Every finitely generated module is the quotient of a finitely generated projective module. I already find a proof of this but it uses tensor functor and flat modules propositions so im looking for a ...
-1
votes
1answer
58 views

Basic short exact sequence exercise in $\mathbb{Z}$-modules [closed]

I was wondering how to properly argue the proof of the following: a) Prove that there is an exact sequence $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_n \to 0$ (consider $f : \mathbb{Z} \to ...
3
votes
4answers
71 views

Proving that P/PJ is a projective right module over R/J

If P is a projective right module over a ring R and J is a two sided ideal of R. Prove that P/PJ is a projective right module over R/J . My idea was trying to proof that " $M$ is an $R$-module ...
1
vote
0answers
28 views

If a direct sum has a projective cover, must the summands have projective covers?

In “Cover of a direct summand” it is asked to show that if a direct sum has a projective cover, and if one of the summands has a projective cover, then so does the other. I gave a solution that works ...
1
vote
1answer
49 views

Cover of a direct summand

Let $L,N$ be $R$-modules. If $L$ and $L \oplus N$ have projective covers, is it true that $N$ admits a projective cover?
0
votes
2answers
43 views

The meaning of functor $M \mapsto \mbox{Hom}_A(P,M)$ being exact

(I'm currently studying Lang's text Algebra and it comes up on the page 137. Lang does not explicitly define this expression.) Is the following understanding correct? The function $M \mapsto ...
3
votes
1answer
52 views

Analogue of Baer criterion for testing projectiveness of modules

In order to test injectivity of a module $M$ it suffices to check if every linear map from an arbitrary ideal extends to the ring or not. Similarly in order to check the flatness of a module $M$ it ...
1
vote
1answer
29 views

The existence of a projective resolution of M from finite rank free modules

Any R-module admits a projective resolution. When the ring is Noetherian, can we show the existence of a projective resolution of any finitely generated R-module by finite rank free R-modules? In ...
1
vote
1answer
29 views

Is there an analogue of Bass-Papp theorem for Projective modules?

The Bass-Papp theorem for injective modules states that If $R$ is a commutative ring such that every direct sum of injective $R$ modules is injective then $R$ is Noetherian. Is there an ...
2
votes
0answers
55 views

Why $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module?

From the fact that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$, how do we conclude that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module?
2
votes
1answer
91 views

Global dimension of $\mathbb Q [x]$

I'm trying to show that the global dimension of $\mathbb Q [x]$ is $1$. I have shown that $D(\mathbb Q [x]) \leq 1$ as follows. One can reduce to the case of showing that ...
1
vote
0answers
28 views

Tensoring and retaining projectiveness

Let $A$be a unital associative ring, If $A$ is not a projective $A$-bimodule, however $A\otimes A$ is a projective $A$-bimodule, can we conclude that $A\otimes A \otimes A$ is also projective?
1
vote
1answer
36 views

Hom($P$, $R$) $\neq 0 $ if $P$ is a nonzero projective left $R$-module (Rotman)

I've found this exercise, number $3.11$ from Introduction to homological algebra. Prove that $\operatorname{Hom}(P, R) \neq 0 $ if $P$ is a nonzero projective left $R$-module. Any hint?
1
vote
1answer
37 views

GCD for multivariable polynomial ring

I'm reading Lectures on Modules and Rings by T. Y. Lam. It's on page 32 of the book, example 2.19A. It reads: (2.19A) Example. Let $k$ be a field. Then in the commutative polynomial ring $R = ...
0
votes
0answers
23 views

Differential operators in the language of modules

I am reading articles about differential operators. Authors try to treat differential systems , by studying the differential operator on a vector space in the language of modules. In this manner ...
1
vote
1answer
45 views

Exact Sequences of R-Modules

In "A Course in Ring Theory by Passman" it is mentioned, "But the kernel of the combined epimorphism $P\rightarrow B\rightarrow C$ is clearly equal to $E$". I don't understand this part. How can the ...
1
vote
1answer
22 views

Projective Dimension and Supremum

Here is a lemma that appears in A Course in Ring Theory by Passman. In the last section of the proof the writer shows that, $\mbox{pd }A_i\leq n\iff \mbox{pd }A\leq n$ and finishes the proof. I don't ...
2
votes
1answer
49 views

Let $k$ be a division ring, then the ring of upper triangular matrixes over $k$ is hereditary

I'm reading Ring Theory by Louis H. Rowen, and he claimed that The ring of upper triangular matrices over a division ring is hereditary (it's on page 196, Example 2.8.13 of the book). I think it ...
3
votes
0answers
61 views

Projective modules over semilocal rings having constant rank are free

I'm starting to study algebraic K-theory by myself and I need a hint how to prove $R$ is a semilocal ring with maximal ideals $\mathfrak m_1,\ldots, \mathfrak m_n$, $P$ is a projective module and ...
0
votes
1answer
78 views

The ideal $I=(3,2+\sqrt {-5})$ is a projective module

Let $R=\mathbb Z[\sqrt{-5}]$ and $I=(3,2+\sqrt {-5})$ be the ideal generated by $3$ and $2+\sqrt{-5}$. I'm trying to prove that $I$ is a projective $R$-module. I'm using the lifting property ...
4
votes
0answers
81 views

Existence of finite projective resolution

The situation I'm considering is quite involved. All rings are noetherian commutative with $1$. All modules are finitely generated. First of all we fix a non reduced local ring $A$ where all zero ...
3
votes
2answers
35 views

Requirements on ring for injective-projectiveness

What requirements could be asked (minimal) of a ring R, so that any module M on R which is injective must also be projective? Is this possible?
7
votes
0answers
146 views

A few questions about a specific ring

My question is kinda long, so please bear with me... And I only need hints to get me started. So, I'm working on the ring $R =\left( \begin{matrix} \mathbb{Z} & \mathbb{Q} \\ 0 & ...
3
votes
1answer
73 views

Question on Nakayama?

In reading a certain proof on the stacks project "http://stacks.math.columbia.edu/tag/00NV", I can't see how Nakayama's lemma is used to make the following conclusion: "Assume M is finitely presented ...
1
vote
0answers
56 views

An invariant submodule of a projective module

Let $R$ be a commutative unital ring and $S$ be a normal subgroup of a finite group $G$. If $P$ is a finitely generated projective left $R[G]$-module then is the submodule of $S$-invariants $$P^S:= ...
3
votes
0answers
58 views

Why $\prod_{i\in I} \Bbb Z$ is not a projective $\Bbb Z$-module? [duplicate]

Let $\Bbb Z$ be the ring of integers and $I$ a non-countable index set. Why $$\prod_{i\in I} \Bbb Z$$ is not a projective $\Bbb Z$-module?
0
votes
1answer
122 views

Prove that $M$ is a free module if and only if $M$ is a projective module over $PID$.

Let $R$ be a principal ideal domain and $M$ a finitely generated $R$ module. Prove that $M$ is a free $R$-module if and only if $M$ is a projective $R$-module. I am quite confused and totally not ...
3
votes
1answer
116 views

Torsion-free and projective modules over a Dedekind domain

Suppose that $A$ is a Dedekind domain (and integral domain). I am trying to prove that if $M$ is a torsion-free $A$-module, then it is projective and vice versa. Suppose that $M$ is projective. Then ...
2
votes
1answer
75 views

T. Y. Lam's definition for the rank of a projective module

This is a concept from Lectures on Modules and Rings of T. Y. Lam. It's on page 34-35 of the book. Throughout this post, $R$ will be used to denote a commutative ring. Let $P$ be a f.g ...
2
votes
1answer
58 views

A noetherian ring is self-injective if and only if all projectives are injective

Let $R$ be a left and right noetherian ring. I try to prove that the following two statements are equivalent. $R$ is injective as a left $R$-module All projective left $R$-modules are injective In ...
1
vote
1answer
57 views

Every projective f. g. module is f. p.

I want to show that if $P$ is a finitely generated (f.g.) projective module then $P$ is finitely presented (f.p.).
4
votes
1answer
258 views

A problem about an $R$-module that is both injective and projective.

Let $R$ be a domain that is not a field, and let $M$ be an $R$-module that is both injective and projective. Prove that $M= \left \{ 0 \right \}$. This is exercise 7.52 of Rotman's Advanced ...
1
vote
2answers
44 views

Direct sum of projective module

Let $\left\{ P_{i}\right\} _{i\in I}$ be a family of $R$-module. I know that if each $P_{i}$ is projective then $\oplus_{i\in I}P_{i}$ is projective. Is the converse true, i.e if $\oplus_{i\in ...
0
votes
1answer
52 views

If $A \cong A^*$, is every projective module also injective?

Suppose $A$ is a finite-dimensional algebra over $k$. Assume further that $A \cong A^* = \text{Hom}(A,k)$ as $A$-modules. My question is: is every finite dimensional projective module over $A$ ...
2
votes
1answer
103 views

Why are projective modules cohomologically trivial?

Let $G$ be a finite group, $H\subset G$ a subgroup, $k$ a commutative ring, $M$ a $kG$-module, $n\in\mathbb{Z}$, and $\hat{H}\,^n(H,M)$ the $n$th Tate cohomology group as defined in this question, ...
1
vote
1answer
146 views

Projective Modules, Annihilators, and Idempotents

Let $Ra$ be the left ideal of a ring $R$ generated by an element $a \in R$. Show that $Ra$ is a projective left $R$-module if and only if the left annihilator of a, $\{r \in R | ra = 0\}$ is of the ...
3
votes
0answers
54 views

Projective Module as a Direct Sum of Left Ideals

I wonder if the following statement is true: Every projective $R$-module is a direct sum of projective left ideals of $R$. Most examples of non-free projective modules I have seen are all left ...
3
votes
1answer
136 views

Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$ and show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module.

1) Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$. 2) Show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module. 1) We know that $\Bbb{Q}$ is an injective $\Bbb{Z}$-module. ...
0
votes
1answer
72 views

Localization at a monic polynomial: Horrocks' theorem

A theorem of Horrocks states that if $P$ is a projective $R[x]$-module ($R$ a local ring) such that $P_S$ is a free $R[x]_S$-module (where $S\subset R[x]$ is the set of all monic polynomials in ...
8
votes
1answer
222 views

A direct product of projective modules which is not projective

I am looking for an elementary example of a family $\{M_\alpha\}_\alpha$ of projective $R$-modules whose direct product is not projective. The simplest example that I know is the $\Bbb{Z}$-modules, ...
2
votes
0answers
54 views

Projective modules are stably isomorphic if localization at a monic polynomial is isomorphic

Let $R$ be a ring and let $P,Q$ be finitely generated projective $R[x]$-modules. Let further $f\in R[x]$ be a monic polynomial. If $P_f\cong Q_f$ (localization on $f$) then $P$ and $Q$ are stably ...
4
votes
1answer
151 views

direct sum of projective modules

I find the following question: Is the direct sum of two projective and not free $R$-modules projective/free? What if $R$ has got no zero divisors? The answer to the first question is very simple, ...
3
votes
2answers
197 views

Is a finitely generated projective module a direct summand of a *finitely generated* free module?

Let $R$ be a (not necessarily commutative) ring and $P$ a finitely generated projective $R$-module. Then there is an $R$-module $N$ such that $P \oplus N$ is free. Can $N$ always be chosen such that ...
0
votes
1answer
49 views

Non-trivial stably free, projective right ideals

I am working on Stably free, projective right ideals (1985) by J.T. Stafford. Trying to get through Theorem 1.2 on page 66 [4]. From the beginning of the paper we are supposed to know: taking two ...
2
votes
2answers
152 views

Example of a simple module which does not occur in the regular module?

Let $K$ be a field and $A$ be a $K$-algebra. I know, if $A$ is artinain algebra, then by Krull-Schmidt Theorem $A$ , as a left regular module, can be written as a direct sum of indecomposable ...