Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

Questions tagged [projective-module]

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

2
votes
1answer
29 views

Projective modules in long exact sequences

Let $A$ be a commutative ring (with unit), and let $(P_i)_i$ be projective $A$-modules sitting in a long exact sequence of $A$-modules: $$0 \longrightarrow P_1 \stackrel{f_1}{\longrightarrow} P_2\...
1
vote
1answer
23 views

Hilbert Syzygy Theorem for non-graded modules

The statement of Hilbert Syzygy Theorem is as follows: Let $R = k[x_1 , \ldots , x_n]$ be a polynomial ring over a field $k$ and $M$ be a finitely generated graded $R$-module. Then $\text{pd }M \leq n$...
2
votes
0answers
53 views

On rings whose $K_0$ has nice properties

Let $R$ be a commutative, reduced ring. It can be seen that $K_0(R) \cong \mathbb Z$ as groups if and only if every finitely generated projective module is stably free. My question is, are there ...
1
vote
1answer
11 views

Projective module which is direct sum of $R^n$

Proposition 4.7.5 Let $P$ be a left $R$-module and $\{u_i\}$ be a generating set for $P$. Then $P$ is projective if and only if there exists $R$-module homomorphisms $\alpha_i:P\rightarrow R$ ...
1
vote
0answers
47 views

Noetherian rings whose prime ideals have projective dimension bounded above

For a module $M$ over a commutative Noetherian ring $R$, let $pd_R (M)$ denote the projective dimension of $M$ as an $R$-module. Now let $R$ be a commutative Noetherian ring such that $\sup \{ pd_R (Q)...
1
vote
1answer
38 views

On a special type of Noetherian regular rings

Let $R$ be a commutative Noetherian ring having the property that for every $R$-module $M$ that has finite projective dimension, every submodule of $M$ also has finite projective dimension. Then $R$ ...
3
votes
1answer
68 views

$M$ is projective iff $M$ is locally free

How to prove this theorem? A finitely presented module is projective iff it is locally free (the localization at every prime ideal is free over the localized ring). This is given right after the ...
-1
votes
0answers
30 views

Confusion About Invertible Modules

According to the Stacks Project, a module is invertible iff it is locally free of rank one.(In the strong sense, not just that the stalks are free).Link So, according to this link, the module is ...
0
votes
1answer
78 views

About symmetric inner product space

How lattice $L = V \otimes _k k[x] $ looks like? $L$ is free over $k[x]$ because $V$ is free over $k$ ; why this statement is true? Also I am trying to prove every unimodular $k[x]$ lattice has form $...
0
votes
1answer
40 views

About Inner product space

So this $\delta$ function is degree function over Euclidean domain A which is simply defined as Now my question is why det L is well defined in first picture and how to get corollary 3.9 and remark ...
1
vote
1answer
39 views

About lattice of finitely generated projective module

Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule $L ⊆ V $is said to be an $A$ -lattice in $V$ ...
0
votes
0answers
34 views

Projective module vs. Free module: dual basis lemma

I am trying to understand definition (characterization) of projective module via dual basis lemma, compared with free module. Let $F$ be a left $R$-module. $F$ is said to be free if (1) ...
0
votes
1answer
51 views

Submodules of modules of finite projective dimension over regular ring

Let $R$ be a regular ring i.e. a commutative Noetherian ring whose localisations at every prime ideal is regular local ring. Then every finitely generated $R$-module has finite projective dimension, ...
4
votes
1answer
58 views

If the dual of a module is finitely generated and projective, can we claim that the module itself is?

Assume that $R$ is a commutative ring and that $M$ is a (left) $R$-module. Assume also that we know for some reason that $M^*:=\mathsf{Hom}_R(M,R)$ is finitely generated and projective as (right) $R$-...
1
vote
1answer
35 views

Projective Resolution of exterior algebra as a module over divided polynomial algebra

Let $\Lambda_\mathbb{Z}[x]$ be an exterior algebra on one generator with $|x|=n$, let $\Gamma_\mathbb{Z}[x]$ be a divided polynomial algebra with $|x_k|= kn$, and suppose that $\Lambda_\mathbb{Z}[x]$ ...
0
votes
0answers
20 views

Projective dimension of module over regular ring is always finite? [duplicate]

Let $R$ be a regular ring, i.e. a commutative Noetherian ring all whose localizations at every prime ideal is a regular local ring. So localization of $R$ at every prime ideal has finite global ...
2
votes
0answers
20 views

When are indecomposable projective modules finitely generated?

What conditions to you need to put on your ring to guarantee that the indecomposable projective modules are all finitely generated? Edit: I was hoping there was some general result for this. If your ...
0
votes
1answer
26 views

projective module, The module not projective

I found a proof " The $\mathbb{Z}$ - module $M=\mathbb{Z} \times\mathbb{Z} \times\dots$ is not projective" in T.Y.Lam - Exercises in Modules and Rings. Proof Suppose that $M$ is projective, so $M$ ...
2
votes
1answer
34 views

Vanishing of $\mathrm{Ext}$ and finite projective dimension

Does it follow from "for all finitely generated $N$, there exists $n_0$ such that for $n\geq n_0, \mathrm{Ext}^n(M,N)=0$" that $M$ has finite projective dimension, for a finitely generated $M$ ? ...
3
votes
1answer
43 views

Why is $\Omega (M)$ a superfluous submodule?

I'm struggling with a result that seems intuitive and that authors don't even bother to prove, so I think there's something stupid that I'm not seeing. Let $M$ be a finitely generated module over $...
5
votes
1answer
78 views

A consequence of Schanuel's lemma

In Carlson's Cohomology and representation theory, the author states Schanuel's lemma, and then derives a consequence that I cannot understand. They define, for a $kG$ module $M$, $\Omega (M)$ to be ...
4
votes
1answer
49 views

If $R$ is a domain which is not a field and $M$ is an $R$-module for which $\bigcap_{r\in R\setminus \{0\}} rM \neq 0$, show $M$ is not projective.

I can't seem to crack this one. I've tried assuming $x \in T := \bigcap_{r \in R\setminus\{0\}} rM$ so that for all nonzero nonunits $r \in R$ there are $x_{n} \in M$ so that $x = r^{n}x_{n}$ for ...
0
votes
1answer
62 views

About group ring and projective module

Let $R$ be commutative ring and $G$ be finite group. Now $A=R[G]$ is the group ring of $G$ with coefficients in $R$. Give an example of right $A$-module which is finitely generated and projective as $...
1
vote
1answer
40 views

Do categories of bimodules have enough projectives?

It is well-known and easy to prove (assuming the axiom of choice) that the category $R\text{-}\mathrm{Mod}$ has enough projectives for any ring $R$: Let $M$ be any $R$-module, and let $P$ be the free ...
2
votes
1answer
25 views

Isomorphism between finitely generated projective modules over algebra of functions

Here is the context: Consider two finitely generated projective modules $M$ and $N$ over the commutative algebra of (smooth or) continous functions on a manifold $\mathcal{M}$ (they are sections of ...
0
votes
0answers
116 views

A question about dual map on modules

Let $R$ be a commutative ring (noetherian if needed), and $P$ a finitely generated projective $R$-module (of constant rank $n$). Let $P^\vee:={\rm Hom}_R(P, R)$ be the dual. For $\varphi\in{\rm Aut}_R(...
0
votes
0answers
18 views

$R$ has an identity and $D$ is a projective unitary right $R$-module then sequence of corresponding tensor products is a short exact sequence

$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is a short exact sequence of left $R$-modules and $D$ a right $R$ module. $0 \rightarrow D \otimes_R A \xrightarrow{1_D \otimes\ f} D \...
0
votes
1answer
38 views

On Noetherian domains whose proper submodules of the fraction field are projective

Let $R$ be a Noetherian domain with fraction field $K$. If every proper $R$-submodule of $K$, containing $R$ (i.e. all $R$-submodules $M$ of $K$ such that $R \subseteq M \subsetneq K$) is projective ...
0
votes
1answer
51 views

About dual of finitely generated projective module

Let $M$ be a finitely generated projective module, $x \in M$ and $x \neq 0$. Then is it true that there is $g \in M^*$ such that $gx \neq 0$? If yes how to prove it? For vector space dual this ...
3
votes
0answers
60 views

cancellation ideals in Noetherian domain

For an ideal $I$ in a commutative ring with unity $R$, let us say $I$ is a cancellation ideal if for ideals $J,K$, $IJ \cong IK \implies J \cong K$. Now let $I$ be a cancellation ideal in a ...
0
votes
1answer
45 views

Projective resolution of short exact sequence: definition in Jacobson

Let $0\rightarrow M'\rightarrow M \rightarrow M''\rightarrow 0$ be a short-exact sequence of modules (over a ring $R$). By a projective resolution of this sequence, we mean (according to Jacobson ...
0
votes
0answers
20 views

How to prove that for $K$ $\mathbb{Z}$-free & $Q$ projective, $K\otimes_{\mathbb{Z}}Q$ is a projective?

Let $K$ be $\mathbb{Z}$-free and $Q$ a projective $\mathbb{Z}G$ module then $K\otimes_{\mathbb{Z}}Q$ is a projective $\mathbb{Z}G$-module. I believe that this follows from the adjoint isomorphism ...
2
votes
1answer
48 views

Finitely generated projective modules are reflexive

How to show $i_P$ is an isomorphism ? We need to show that it is injective and surjective.
5
votes
2answers
121 views

Every $\mathbb{Z}/6\mathbb{Z}$-module is projective

I have to prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective. I've found already this question Prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective and injective. Find a $\...
0
votes
0answers
45 views

Calculation of Tor for module with trivial action

Let $S$ be a ring of the form $\mathbb{Z}/p[x] \otimes_{\mathbb{Z}/p}E(y)$. Where $\mathbb{Z}/p[x]$ is the polynomial algebra and $E(y)$ is the exterior algebra over $\mathbb{Z}/p$. Consider $\bar{S},...
0
votes
1answer
20 views

If possible, construct a simple module that is not projective.

Suppose that $M$ is a simple (or irreducible) $R$-module. Does this imply that $M$ is projective? Vector spaces are all projective since they are free, so no counter example can be gotten from there....
1
vote
1answer
56 views

Noetherian domain whose fraction field is such that some specific proper submodules are projective

Let $R$ be a Noetherian domain (which is not a field) with fraction field $K$. Suppose every proper $R$-submodule of $K$ of the form $R[1/a]$, where $a\in R$, is projective as an $R$-module. Then, I ...
1
vote
1answer
90 views

Projective cover of modules

Give me a hand please, how can i prove this sentence. "Show $\mathbb{Z}_{2}$ doesn't have projective cover as $\mathbb{Z}$-module."
1
vote
1answer
48 views

Finitely generated free module is projective.

Call a $R$-module projective if every short exact sequence $0 \to A\stackrel{f} \to B\stackrel{g} \to C \to 0$ of $R$-modules splits. Call a short exact sequence as above split, if it admits a ...
0
votes
0answers
55 views

Generalized Schanuel Lemma

This is on page 128, ex 3.15, of Rotman's AIHA, (Schanuel) Let $B$ be a left $R$-module over some ring $R$ consider two exact sequences, $$ 0 \rightarrow K \rightarrow P_n \rightarrow \...
1
vote
1answer
36 views

If P is a Projective R Module and Q is an Injective R module then $P \otimes_{R} Q$ is Injective

I am doing some basics on Protective, flat and Injective but I have no idea how to proceed for this one. Any help is appreciated!
0
votes
2answers
68 views

length of minimal projective resolution and projective dimension

Let $K$ be a field, $A$ be a $K$-algebra, and $M$ be a finite generated $A$-module. How do you show that the length of a minimal projective resolution of $M$ is the projective dimension of $M$? (This ...
1
vote
0answers
71 views

Divisible module over an integral domain is not projective?

I was reading the proof of the fact that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module. And I was wondering if we could say that any divisible module over an integral domain is not projective?
3
votes
0answers
23 views

From root and weight lattices of SU(N) to $\theta$-functions as sections of a line bundle and $CP$-space

I have troubles to digest the following messages/discussions in the following work in p.10-12; Which construct a map from the moduli space of flat connections $M_{\rm flat}=\mathbb{E} / {\mathfrak S}...
0
votes
1answer
25 views

$P$ direct summand of a free module implies $P$ is projective

I've looked at several posts here and proofs elsewhere and it is not clicking in my brain. Let $R$ a ring and $P$ be a direct summand of a free module. That is, there exists a (left) $R$-module $Q$ ...
2
votes
2answers
83 views

Intuition for definition of projective module

A free module is rather intuitive: there exists a subset satisfying suitable properties which may act as a basis for elements. Taking as example an $\mathbb R$-module, or a vector space, we see "free" ...
1
vote
0answers
39 views

commutative Noetherian ring whose every maximal ideal is projective

Let $R$ be a commutative Noetherian ring. If every maximal ideal of $R$ is projective as an $R$-module, then is $R$ hereditary?
2
votes
0answers
64 views

$\mathbb{T}^4/\mathbb{Z}_2$, $\mathbb{T}^2/\mathbb{Z}_2$, and $\mathbb{CP}^1$ [closed]

I was stuck by reading this figure: It looks that $\mathbb{T}^4/\mathbb{Z}_2$, $\mathbb{T}^2/\mathbb{Z}_2$, and $\mathbb{CP}^1$ are somehow related. Are there some easier explanations from math ...
1
vote
0answers
85 views

Stable module category and Syzygy functor

I'm trying to understand the construction of the functor $\Omega:{}_{R}\underline{\mathfrak{M}}\longrightarrow {}_{R}\underline{\mathfrak{M}}$ of first syzygia, but I don't understand how it's define: ...
6
votes
1answer
135 views

Ideal $I$ in a Noetherian ring such that $I+(b)$ and $I+(a)$ are projective, where $ab\in I$

Let $R$ be a Noetherian ring and $I$ be a proper ideal of $R$ and let $a,b\in R$ be such that $ab\in I$ and $I+Ra$ and $I+Rb$ are projective as $R$-modules. Then is the ideal $I$ projective as an $R$...