Questions tagged [projective-module]

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

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Is $C$ projective as $C[x]$ module?

I'm asking myself if $C = C[X]/(x)$ is a projective/flat $C[X]$ module. I found on the internet that this isn't the case, but on the other hand we have $ C[X] = C \oplus xC[X]$, so $C$ is a direct ...
2 votes
2 answers
46 views

Prove that $\mathbb{Z}/ 3\mathbb{Z}$ is a projective $\mathbb{Z} / 6\mathbb{Z}$ module which is not free.

I am a student of a masters course and this question was asked in my quiz of commutative algebra. Question: Prove that $\mathbb{Z}/ 3\mathbb{Z}$ is a projective $\mathbb{Z} / 6\mathbb{Z}$ module ...
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1 vote
1 answer
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Understanding key step of dimension shifting proof in homological algebra

I'm trying to do the following exercise from Weibel's Introduction to homological algebra regarding 'dimension shifting' Exercise 2.4.3 : If $0\to M\to P\to A \to 0$ is exact with $P$ projective (or $...
1 vote
1 answer
34 views

In the definition of a projective object, when do we NOT lose generality from restricting the potential non-epi to an epi?

From Wikipedia: An object $P$ in a category $\mathcal{C}$ is projective if for any epimorphism $e:E\twoheadrightarrow X$ and morphism $f:P\to X$, there is a morphism $\overline{f}:P\to E$ such that $e\...
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Functors and projective covers

I'm looking to understand how a covariant functor that is an equivalences of categories preserves projective covers, and how a contravariant functor that is a dual equivalence of categories maps ...
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5 votes
1 answer
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The functor $\mathrm{Hom}(A,-)$ cannot commute with arbitrary direct sums for infinitely generated projective module $A$

It is easy to see the functor $\mathrm{Hom}(A,-)$ commutes with every arbitrary direct sum (i.e. $\mathrm{Hom} (A,\oplus_{i\in I} N_i)=\oplus_{i\in I}\mathrm{Hom}(A,N_i)$) for finitely generated ...
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1 vote
1 answer
40 views

Examples of rings by their relation to projective coverings

A projective covering of an $R$-module $M$ is an epimorphism $\pi:P\rightarrow M$ s.t. $P$ is a projective $R$-module and $\textrm{Ker}(\pi)$ is co-essential in $P.$ The existence theorem for ...
0 votes
1 answer
97 views

$A/I$ is a projective $A$-module iff $ I$ is a principal ideal generated by an idempotent element

The following question was asked in my assignment of modules and I could not solve this question despite thinking a lot. Question: Show that $A/I$ is a projective $A$-module if and only if $I$ is a ...
0 votes
1 answer
33 views

Existence of non-trivial homomorphism from projective module to ring

Let $R$ be a commutative ring and let $P$ be a nonzero projective $R$-module. I want to show that there exists a non-trivial homomorphism from $P$ to $R$. I don't see how can I start. Can someone give ...
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3 votes
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Description of projective $\mathrm{SL}_{2}(\mathbb{Z})$-modules

I am working my way through Ken Browns book on the cohomology of groups, and in particular chapters 8 and 9 on finiteness conditions, and Euler characteristics. Most of the concepts in chapter 9 (such ...
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58 views

Projective resolutions of modules over a valuation ring

Let $K$ be the field of Hahn series in an indeterminate $t$ with exponents in $\mathbb{R}$, coefficients in $\mathbb{F}_2$ and valuation $v$. For each $q\in\mathbb{R}$, we set $$I_q:=\{a\in K:v(a)\geq ...
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Every nonzero projective module has a maximal submodule

Let $P \neq 0$ be a projective right module over a ring $R$ with unity. I need to prove that $P$ has a maximal submodule (This is equivalent to saying that the radical of $P$ is a proper submodule of $...
0 votes
0 answers
48 views

$M$ projective and finitely generated implies $Hom_R(M,R)$ is projective and finitely generated

I'm trying to prove that if $M$ is a left $R$-module projective and finitely generated, then $Hom_R(M,R)$ is a right $R$-module projective and finitely generated. I've seen comments where they use ...
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34 views

Nil(R) is an exact category

The definition of an exact category as I have read is a pair ($\mathcal{C}, \mathcal{E}$) is an exact category if the additive category $\mathcal{C}$ is embedded as a full subcategory inside an ...
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2 votes
0 answers
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Syzygy in projective resolution and thick closure

Let $R$ be a Commutative Noetherian ring, and let $\text{mod } R$ denote the abelian category of finitely generated $R$-module. Consider the bounded derived category $D^b(\text{mod } R) $ which is a ...
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1 vote
1 answer
49 views

Is trace equal to 1 equivalent to free on one generator?

Let $R$ be a commutative ring and $M$ a f.g. projective module. Is it true that if the trace of the identity map of $M$ is equal to 1, then $M \cong R$?
1 vote
1 answer
81 views

linear map between finitely generated free modules (over local ring) which become injective after tensoring with the residue field

Let $F,G$ be finitely generated free modules over a Noetherian local ring $(R,\mathfrak m)$. Let $f: F\to G$ be an $R$-linear map such that the induced map $\bar f: F/\mathfrak m F \to G/\mathfrak m G$...
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1 vote
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Modules and homomorphisms

Let $R$ be a unital ring not necessarily commutative and $f:<a>\to<b>$ is a $R$-homomorphism of cyclic $R$-modules. Let $X=\lbrace (r,s)\in R^2\text{ such that } r.a\neq 0 \text{ and} \...
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1 vote
1 answer
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$\text{Hom}_k(P,P)$ is isomophic to $P\otimes_k \text{Hom}_k(P,k)$

Let $P$ be a finitely generated projective right module over $k$. Then, $$\text{Hom}_k(P,P)\cong P \otimes_k \text{Hom}_k(P,k).$$ I was able to show the congruence assuming $P$ is a vector space over $...
1 vote
1 answer
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The proof of Kaplansky's threorem

I am reading "Commutative Ring theory" written by Matsumura. I have a question about the proof of Kaplansky's theorem. In Lemma 1, we construct a well-ordered family $\{F_{\alpha}\}$ such ...
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10 votes
1 answer
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Can we classify all finitely generated projective modules over $k[x,y,x^{-1},y^{-1}]$?

Let $k$ be a field and we consider the ring $R=k[x,y,x^{-1},y^{-1}]$. Can we classify all finitely generated projective modules over $R$? In particular, are there non-free examples? I considered the ...
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2 votes
1 answer
60 views

Can a projective module be singular?

Let $R$ be a ring with unity, and $M$ be a cyclic projective $R$-module. I know that $M$ can not be singular. My question is, if we remove the cyclic condition, will the conclusion hold?
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1 vote
2 answers
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Isomorphism in Projective modules over the algebraic closure of a finite field Fp

Let $F \otimes Q \cong F\otimes R$. Does it imply $Q\cong R$? where $R$ and $Q$ are modules over a field containing $\mathbb{F}_p$ and F is the algebraic closure of $\mathbb{F}_p$.
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1 vote
1 answer
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does the flat pre-cover need to exist for a left module?

All modules do have the flat cover by a result from the year 2001. What can be a ring $R$ and a left $R$-Module $M$ which doesn't have (a/the) flat pre-cover ? How can I construct such $R$ and left ...
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1 vote
0 answers
31 views

Projective resolution of a $K[t]$-module.

Let $K$ be some field. I want to show that every indecomposable right $K[t]$-module $M$ is has a projective resolution of the form $$0 \to P_1 \to P_0 \to M \to 0.$$ To that end, I have a hint: Such a ...
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1 vote
0 answers
50 views

When the thick closure (in bounded derived category) of every finitely generated module contains a non-exact perfect complex

Let $R$ be a commutative Noetherian ring and $\mod R$ be the (abelian) category of finitely generated $R$-modules. Let $\mathcal D^b(\mod R)$ be the bounded derived category of $\mod R$. Each finitely ...
1 vote
0 answers
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Elements of $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i\geq 0}\mathbb{Z}, \mathbb{Z})$ vanish on almost all elements of "standard basis" $\mathbb{e}_{n}$

I've been struggling with the following exerciese: Let $f\in\mathrm{Hom}_{\mathbb{Z}}(\prod_{i\geq 0}\mathbb{Z}, > \mathbb{Z})$, where $\prod_{i\geq 0}\mathbb{Z}$ denotes the $\mathbb{Z}$-module ...
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2 votes
2 answers
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Rank and restriction of scalars

Let $R \hookrightarrow S$ be commutative rings. Suppose that $M$ is a finitely generated projective $S$-module. Let $f : \text{Spec}(S) \rightarrow \text{Spec}(R)$. We have locally constant rank ...
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2 votes
0 answers
42 views

About vanishing of Tor of a morphism over a perfect ring?

Let $R$ be an associative ring with identity. An ideal $I$ of $R$ is said to be T-nilpotent if for every left $R$-module $M$, $IM=M$ implies $M=0$ equivalently for every left $R$-module $M$, $Hom_{R}(...
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0 answers
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$M$ is flat iff $\operatorname{Tor}(M,R/J(R))=0$?

A commutative ring $R$ is said to be perfect if every flat $R$-module is projective. $J(R)$ denote the radical of Jacobson of $R$ wich is the intersection of all maximal ideals of $R$. I want to know ...
0 votes
0 answers
34 views

homotopies between cofibrant resolutions

Let $X$ be a finite-dimensional noetherian separated scheme, let $\mathcal{U}=\{U_i\}$ be an affine cover and $\mathcal{N}$ the nerve of this cover. For any coherent sheaf $\mathcal{F}$ on $X$, we ...
1 vote
0 answers
32 views

Finitness conditions under change of rings?

Let $R \to S$ be a homomorphism of commutative rings, such that $S$ is finitely generated and projective $R$-module. Proof that if $A$ is a finitely presented $S$-module then $A$ is a finitely ...
0 votes
1 answer
136 views

If every homomorphic image of an injective module is also injective, then every submodule of a projective module is projective

Let $R$ be a ring with $1$. All modules considered in this problem are unitary right $R$-module. Assume that every homomorphic image of an injective module is also injective. I need to prove that ...
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0 answers
80 views

If $R$ is a right hereditary ring, then any submodule of a right projective $R$-module is again projective.

Recall that a ring $R$ is called right hereditary if every right right ideal $I\subset R$ is projective as a right $R$-module. I need to prove that if $M_R$ is projective module and $N\leq M$ is any ...
0 votes
1 answer
55 views

Example of invertible modules?

I am trying to find an example of a non-trivial invertible module (let's say over $\mathbb Z$). This seems to be very simple, but after trying and searching around, I do not find any examples. (Many ...
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3 votes
1 answer
106 views

Projective dimension of $K[x,y]$ over $K[xy]$

Let $K$ be a field and $K[x,y]$ the polynomial ring in two variables $x$ and $y$ over $K$. Let $R = K[xy]$ be the subring generated as a $K$-algebra by the monomial $xy$. My question is: What is the ...
0 votes
1 answer
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Unimodular element of a module

Let $A$ be a noetherian ring and $M$ an $A$-module. An element $z \in M$ is said to be unimodular if $Az$ is a direct summand of $M$ and $Ann(z) = \{r \in A | rz=0\} = 0$. The order ideal of $M$ is: $...
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1 answer
60 views

A projective module over a finite direct product of fields

Is a projective module $P$ over a ring $R$ which is a finite direct product of fields free?
1 vote
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34 views

Lifting Property of flips

I am trying to understand the concept of flips before learning the Bass Cancellation theorem. Let $R$ be a Noetherian ring and $P$ be a projective $R$ module. Let $p,q \in P$ , $\phi \in Hom(P,R)$ ...
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74 views

Question On Serre's Splitting Theorem

I am learning the splitting theorem from the book F. Ischebeck and Ravi Rao. The statement is as follows: Let $A$ be a commutative Noetherian ring of finite Krull dimension. Let $P$ be a finitely ...
1 vote
1 answer
46 views

$\{P\ \text{is a prime of} \ R|\operatorname{rank} M_P=r\}$ is open in $\operatorname{Spec}R$

Let $R$ be a commutative Noetherian ring with identity. $M$ is a finitely generated projective $R$-module. Then for any $r$, the set $\{P\ \text{is a prime of} \ R|\operatorname{rank} M_P=r\}$ is an ...
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Injective projective modules over left artinian ring

When I read Rings and Categories of Modules written by Frank W.Anderson and Kent R.Fuller, I can't understand the proof of Theorem 31.3 (Page 338) Let $R$ be a left or right artinian ring with $J=J(R)...
2 votes
1 answer
43 views

Direct limit of n-presented modules?

A module $M$ is said to be $n$-presented If there exist an exact sequence $$F_{n}\to F_{n-1}\to \cdots \to F_{1}\to F_{0}\to M$$ with each $F_{i}$ is free finitely generated. For example $M$ is $0$-...
1 vote
0 answers
73 views

Prove that if $\mathcal{A}$ has enough projectives, then so does $Ch(\mathcal{A})$

This is exercise 2.2.2 in Weibel's AIHA. We already know that a chain complex $P_{\bullet}$ is projective in $Ch(\mathcal{A})$ iff it is a split exact complex of projectives. Here's my proof, but it ...
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2 votes
0 answers
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Why does tensor product satisfy pushout-product axiom?

Example 11.4 in this paper claims that the tensor product of chain complexes of bimodules (over not-necessarily-commutative rings) satisfies the pushout-product axiom (the first condition of a Quillen ...
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To show 'dual' of a Projective is projective

I am reading Chapter IV, section 2 in Assem's book Elements of the Representation Theory of Associative Algebras and I am stuck at a claim: Let $A$ be a finite dimensional algebra over $\mathbb{C}$. ...
1 vote
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Computing derived tensor product of bimodules by resolving only one argument?

If $R$ is a ring, $M$ is a right $R$-module, and $N$ is a left $R$-module, the derived tensor product $M \otimes_R^{\mathbf{L}} N$ is computed by choosing projective resolutions $P_* \to M$ and $Q_* \...
1 vote
0 answers
66 views

About exactness of $Hom$ functor?

We know that a module $P$ is projective if and only if the functor $Hom(P,-):R-Mod \to Ab$ is exact, i.e it preserves epimorphisms: If $\alpha: M \to N$ is an epimorphism of modules then $Hom(P,\alpha)...
1 vote
1 answer
52 views

$0\to \text{Hom}(P_0,G)\to \text{Hom}(P_1,G)\to \text{Hom}(P_2,G)\to\dots$ is an exact sequence. [duplicate]

I am reading ''Algebraic number theory'' by Cassels and Fröhlich and in the chapter IV.4 it says the following: If $\dots\to P_2\to P_1\to P_0\to\mathbb{Z}\to 0$ is a projective resolution and $G$ is ...
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2 votes
0 answers
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Phantom morphisms and Tor functor?

$\newcommand{\Tor}{\operatorname{Tor}}$In the category of R-modules, a morphism $f:M\to N$ is called a phantom morphism if for every finitely presented module $F$ and every morphism $g:F\to M$, $fg$ ...

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