1
vote
0answers
26 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
23 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
24 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
11 views

A Ring with all Cyclic R-modules has Finite Projective Dimension but Infinite Global Dimension.

Can anybody give me an example of a ring $R$ with the property that each cyclic $R$-module has finite projective dimension even though the Global Dimension of $R$ is equal in infinity?
0
votes
0answers
20 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 ...
0
votes
0answers
16 views

Exact Sequences of R-Modules

Here's a lemma in A Course in Ring Theory by Passman. In the proof it is mentioned, "But the kernel of the combined epimorphism $P\rightarrow B\rightarrow C$ is clearly equal to $E$". I don't ...
1
vote
1answer
15 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
45 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 ...
2
votes
0answers
48 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
67 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
67 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 ...
2
votes
2answers
26 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?
0
votes
0answers
30 views

Quotient of projective indecomposable modules by their radicals

Assume $\Lambda$ is an artin algebra and let $P$ be an indecomposable finitely generated projective $\Lambda$-module. Let $J(P)$ denote the Jacobson radical of $P$ (which, in this setting, is the ...
6
votes
0answers
137 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
68 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
42 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
54 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
63 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 ...
2
votes
1answer
57 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
50 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
55 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.).
3
votes
1answer
181 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
43 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
47 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
71 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
106 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
45 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 ...
1
vote
1answer
87 views

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

1) Prove that $Hom_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$ 2) Show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module. 1) We know that $\Bbb{Q}$ is an injective $\Bbb{Z}$-module. This implies ...
0
votes
1answer
69 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
132 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
48 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
135 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
155 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
139 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 ...
1
vote
0answers
47 views

Rejects and injectives

Let $A$ be any ring and consider modules on the left. It is well known that the trace $Tr(M,A)$ is a two-sided ideal of $A$. If $A$ is a unitary ring then: $Tr(P,A)P=P$, for $P$ projective; ...
1
vote
1answer
128 views

Projective cover of a simple module and indecomposable factor modules

Let $P$ a non-zero projective module. Prove that $P$ is the projective cover of a simple module if and only if every non-zero factor module of $P$ is indecomposable. Thank you.
5
votes
1answer
158 views

Modules which are isomorphic to their tensor product.

Suppose that we have a commutative ring $R$. I am interested in finding the (finitely generated and projective, if you want) $R$-modules $M,$ such that $M\cong M\otimes_R M$ as $R$-modules. I know ...
7
votes
1answer
135 views

Question on Projective Dimensions

$\require{AMScd}$I have a question regarding a claim in A first course of homological algebra by Northcott. I think it's very easy, since the author didn't provide a proof, and just kind of claimed ...
8
votes
2answers
253 views

Finitely generated flat modules that are not projective

Over left noetherian rings and over semiperfect rings, every finitely generated flat module is projective. What are some examples of finitely generated flat modules that are not projective? Compare ...
2
votes
2answers
37 views

Partial cycles in projective resolutions of square-free algebra

Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand? I suspect not, but have not ...
7
votes
1answer
116 views

Specific projective dimension of a module over bound quiver

Suppose $K$ is an algebraically closed field, and $A$ is the algebra presented by the quiver $$\require{AMScd} \begin{CD} 1 @>>> 2\\ @V{}VV @V{}VV \\ 3 @>>> 4 @>>> 5 ...
4
votes
2answers
209 views

Injective and Projective module

Ok, this problem is driving me nuts. At first, I thought I did it. But when reading another textbook (having a similar proposition, they (the problem in my textbook, and the proposition in the other ...
0
votes
1answer
28 views

A module which is not singular

Suppose that M is a projective R-module and that it's simple, isomorphic to $R/I$ where $I$ is a maximal left ideal of $R$ such that $I$ is not a direct summand of $R$. How to get to a contradiction?
2
votes
2answers
58 views

Exact sequences and finitely generated projective modules

Let $R$ be a ring and $A,B,C$ be $R$-modules in an exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0.$$ Submodules or quotients of finitely generated projective modules are not ...
2
votes
1answer
128 views

Resolutions of bimodules as $R^e$-modules.

Let $k$ be a commutative ring, let $R$ be a $k$-algebra, a $R$-Bimodule $M$ over $R$ is a $k$-module with two actions of $R$ on $M$, on the left and on the right, the classical example of this being ...
7
votes
1answer
393 views

Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?

Why isn't an infinite direct product of copies of $\Bbb Z$ a free module? Actually I was asked to show that it's not projective, but as $\Bbb{Z}$ is a PID, so it suffices to show it's not free. ...
3
votes
1answer
124 views

vector bundles on the affine line over a PID

Let $R$ be a PID. Is every finitely generated projective $R[T]$-module free? In other words, is every vector bundle on $\mathbb{A}^1_R$ trivial? For $R=k[X]$ this is true by the Theorem of ...
4
votes
0answers
73 views

Separability of finitely generated projectives over commutative ring

A class $\mathcal{C}$ of $R$-modules is called -separative if $A \oplus A \simeq A \oplus B \simeq B \oplus B$ implies $A \simeq B$ for each $A,B \in \mathcal{C}$ -cancelative if $A \oplus C \simeq ...
1
vote
1answer
41 views

Cancellation law of morphisms?

The title is probably misleading, but I wasn't quite sure how to boil my question down to one line. My problem is this: Assume I have two projective $R$-modules, $P$ and $Q$, and epimorphisms ...