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

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$R$ integral domain, $P$ projective and injective module $\implies P=0$ or $R$ is field of fractions [duplicate]

I'm having difficulties proving the following: Let $R$ be an integral domain and $P$ a projective and injective $R$-module. Show that $P=0$ or $R=Q(R)$, where $Q(R)$ denotes the field of fractions of ...
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Functor Derived, homological algebra, projective modules

Anyone help me with this question Question: Show that if M is module projective, then $L_0FM = FM$ and $L_nFM =0$ for n > 0. p.s. exercise 1, page 346 Basic algebra, vol. 2, Jacobson Thanks in ...
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1answer
30 views

Is $\Bbb Z_2$ free?

As a counter example for a projective module which is not free my instructor gave this one: $\Bbb Z_6=\Bbb Z_2\oplus\Bbb Z_3$ $\Bbb Z_6$ projective, so $\Bbb Z_2$ and $\Bbb Z_3$ are projective but ...
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1answer
22 views

Projective Module equivalence proof [closed]

Given an $A$-module $P$. How can I show that if there exists a module $Q$ such that $P \oplus Q$ is free then the functor $\operatorname{Hom}(P,\cdot)$ is exact?
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1answer
36 views

Projective Module equivalence [closed]

I'm trying to prove that given an $A-module$: $P$ The functor $Hom(P,-)$ is exact $\implies$ $\exists Q (A-module): P\oplus Q$ is free Can anyone guide me with some hints? It's the first time I'...
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1answer
24 views

Show that $\mathbb{Z}_2$ is not a free $\mathbb{Z}_6$-module

Let $ \Lambda = \mathbb{Z}_6 $,the ring of integers modulo $ 6 $.Since $ \mathbb{Z}_6 = \mathbb{Z}_2 \oplus \mathbb{Z}_3 $ as a $\mathbb{Z}_6 $-module,then $\mathbb{Z}_2$ as well as $\mathbb{Z}_3$ are ...
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20 views

Direct summand of module

Let $S$ be a commutative (associative) ring with $1$, which is an extension of a commutative ring $R$ (with same $1$). Suppose that $S$, as an $R$-module, is finitely generated projetive and faithful. ...
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1answer
56 views

Projective module with non-zero annihilator [closed]

Let $M$ be a projective module. Suppose $\operatorname{Ann}_{R} \left(M \right) \neq 0$, where $\operatorname{Ann}_{R} \left( M \right) =\{r\in R : mr = 0, \ \forall m \in M \}$. Then there exists an ...
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1answer
31 views

If every cyclic $R$-module is projective, then $R$ is semisimple.

If every cyclic $R$-module is projective, then $R$ is semisimple. I know how to prove this if the definition of cyclic module is $M=Rm$ where $m$ is an element of $M$. The problem is, we defined in ...
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1answer
49 views

Lifting homomorphism when module is direct summand of free module [closed]

Let $N,M,M',N'$ be modules such that $F = N \oplus N'$ is a free module. Let further $f: M \rightarrow M''$ be a surjective homomorphism. I would like to show that for any homomorphism $\phi: N \...
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1answer
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Natural action of $\mathbb{Z}G$ on $\mathbb{Z}$?

I'm studying projective modules and I'm having problem coming up with (or understanding) examples of non-free projective modules. I got that when a ring is a direct sum $R = A \oplus B$, both $A$ and $...
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212 views

Computing (the ring structure of) $\mathrm{Ext}^\bullet_R(k,k)$ for $R=k[x]/(x^2)$

Let $k$ be some field (say of characteristic zero, if it matters) and define $$R=k[x]/(x^2).$$ I want to compute $$\mathrm{Ext}^\bullet_R(k,k)$$ and, in particular, the ring structure on it (though I ...
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20 views

Projections and projective modules of $C(X)$

This is a followup to this question I made yesterday (disclaimer: I'm new here and I'm not sure if asking a new but related question is the correct procedure). If $X$ is a connected, compact, ...
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1answer
41 views

Projective Dimension and Schanuel's Lemma

Let $R$ be a ring and $M$ a (say, left) $R$-module of projective dimension $n$. According to Noncommutative Noetherian Rings, any projective resolution of $M$ can be terminated at length $n$, and this ...
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33 views

Any characterization for commutative rings over which “projective modules” equal “free modules”?

As far as I know, over any PID, an polynomial rings over a field, or an local ring, projective modules are always free. This kind of results make me curious about if there are any overall ...
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1answer
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If $P/I_kP$ are finitely-generated, is it true that $P/IP$ is finitely-generated where $I=\bigcap I_k$?

I'm looking into some old results on "big projectives'', and trying to understand some steps. Assume that $R$ is a (commutative) ring and $I_1,\ldots,I_n$ are ideals. Let $I=I_1\cap\cdots\cap I_n$ be ...
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1answer
106 views

Direct proof that infinite product of copies of $\mathbb{Z}$ is not projective

It is well-known that the abelian group $$A = \prod_{n=1}^\infty \mathbb{Z}$$ is not free (see, for example this MO question), and that over a PID being free is equivalent to being projective (see ...
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1answer
54 views

Composition factors of injective indecomposable and projective indecomposable modules

Let $A$ be a finite-dimensional algebra over an arbitrary field $K$. Let $L_1$ and $L_2$ be simple modules such that $L_1 \not \cong L_2$. Let the $A$-module $Q_1$ be the injective hull of $L_1$, ...
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Let $e \in R$ be an idempotent. Show $Re$ is a projective $R$-module.

Suppose $e$ is an idempotent element of a unital ring $R$. I'm trying to show $Re$ is a projective left $R$-module. I've seen a few answers out there about showing $Re \oplus R(1-e) = R$ to finish the ...
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1answer
25 views

on the splitting of a presentation

Let $P$ be an $R-$module and let $F$ be a free module (say $F$ isomorphic to $R^n$). If $P$ is a summand of $F$, clearly there is a presentation (namely a surjective $R-$homomorphism) $p:F\to P$. ...
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2answers
49 views

Is $\mathbb{R}$ a projective $\mathbb{Z}$-module?

I found out that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module, because it can not be a direct summand of a free $\mathbb{Z}$-module. What about $\mathbb{R}$, is $\mathbb{R}$ a projective $\...
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Analogue of the trivial extension for higher Ext.

I've been doing some homological algebra and some work on showing some extensions are equivalent, and a thought just came to me, which is that I didn't know how to write down what the analogue of the ...
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1answer
58 views

How to prove the following sequence is exact?

Let R be a ring and $F',F,F'',G',G,G''$ left R-modules. Assume we are given R-module homomorphism $i:F'\to F,p:G'\to G,p':G\to G''$ and $a:F'\to G',b:F\to G,c:F''\to G''$ such that the following ...
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1answer
31 views

definitition of projective resolution of $R$-modules (with homology)

Let $R$ be a commutative ring with unit $1_R$, $M$ be a $R-$module. I have a small question about different definitions of projective resolutions of $M$ (and I'm confused with the degrees of the ...
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1answer
23 views

Are torsion-free modules over principal ideal domains/Dedekind domains projective

An exercise in "Commutative algebra with a view towards Algebraic geometry" by Eisenbud states that a torsion-free module over a Dedekind domain is a projective module (see page $484$, Exercise $19.6$)...
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210 views

Grouping Problem

Suppose there are 9 strangers. We will assign them into 3 groups and each group has exactly 3 people. For each grouping, the strangers who were assigned into the same group will get to know each other ...
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1answer
20 views

On the equivalence condition of p.p. rings

A ring is known to be left p.p. if every principal left ideal is projective. It is well known that this condition is equivalent to the fact that every annihilator of each element is generated by an ...
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If an R-module P is free and A and B are direct summands of P then A∩B is isomorphic to a direct summand of P? is it true? I could not prove it?

If an R-module, P, is free and A and B are direct summands of P, then $A\cap B$ is isomorphic to a direct summand of P. I couldn't prove this proposition on my own. Is it true?
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Left vs right projective resolutions and homology of monoids

Let me use the ad hoc notation $\mathbb Z^l$ and $\mathbb Z^r$ to distinguish between left and right modules. These are trivial modules. The homology of a (discrete) finite monoid $M$ with ...
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2answers
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Proving module homomorphism has right inverse [duplicate]

I have met this as part of a problem in module theory which states Let $f : M \to U $ be a surjective module homomorphism over ring $R$ where $M$ is finitely generated and $U$ is free. How would I ...
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1answer
87 views

Prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective and injective. Find a $\mathbb{Z}/4\mathbb{Z}$-module that is neither.

I want to show that every $\mathbb{Z}/6\mathbb{Z}$-module is a direct sum of projective modules. I know that $\mathbb{Z}/6\mathbb{Z}$ is the direct sum of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\...
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47 views

projective resolution of finitely generated modules

I am in the condition where I have a noetherian ring $R$ of finite global dimension. Consider the category of finitely generated (right) modules over $R$. Then I want to show that every module admits ...
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2answers
56 views

Prove that $\operatorname{Ext}^{d+1}(A, B)\cong \operatorname{Ext}^1(M_d,B)$

So, given a resolution, with $P_{i}$ projective modules: $$0\longrightarrow M_d\longrightarrow P_{d-1} \longrightarrow \cdots \longrightarrow P_0 \longrightarrow A\longrightarrow 0,$$ I'm trying to ...
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Projective objects of chain category

I am tempted to think that the projective objects in the chain category $\text{Ch}(\mathcal C)$ for $\mathcal C$ abelian are exactly the complexes $P_\bullet$ for which each $P_i$ is projective. Is ...
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2answers
110 views

Can you prove that over a division ring , every module is both projective and injective?

I have problem with this : over a division ring or a ring with identity every module is both projective and injective . Can you help me? thank you .
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$\text{Hom}$ to a projective $D[G]$-module for a complete DVR $D$

Suppose you have a complete DVR $D$ and a finite group $G$ with $D[G]$-modules $A$ and $B$. Does $B$ being projective imply that $\text{Hom}_{D[G]}(A,B)$ is $D$-free? Or should it be $A$ that's ...
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1answer
81 views

$k[x]\otimes k[x]$ as a right $k[x]$-module

Let $k$ be a commutative ring. Consider the ring map $\varphi:k[x]\to k[x]\otimes_k k[x]$ given by $\varphi(x)=x\otimes 1-1\otimes x$. Now consider $k[x]\otimes_k k[x]$ as a right module over itself. ...
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28 views

Projective covers for simple Lie algebras in characteristic zero

I want to use the projective cover for the trivial module over a finite-dimensional simple Lie algebra in characteristic zero. Is there a reference that I can quote to assert its existence?
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1answer
74 views

When is the localization of a commutative ring a finitely generated projective module?

Let $R$ be a commutative ring and $M$ an $R$-module. The tensor product $(-)\otimes M$ has a left adjoint $(-)\otimes M^\ast$ for $M^\ast =\mathsf{hom}(M,R)$ iff $M$ is finitely generated projective. ...
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$P\cong P^\ast$ iff $P$ is a f.g projective module?

Is it true that for a noncommutative $R$, a module $P$ is f.g projective iff $\mathsf{hom}(P,R)=P^\ast \cong P$? Here's what I thought of as a proof: Since $(-)^\ast$ is additive, it preserves ...
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27 views

Verification of Ext groups and projective resolution for S3 over F3

So I've been looking at Ext groups of irreducible representations of $S_3$ over $\mathbb{F_3}$. Specifically, I'm doing a project where I'll be looking at extensions themselves, so am really only ...
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1answer
49 views

Are finitely generated projective models of an algebraic theory always finitely presented?

I know that for modules over rings, a finitely generated projective module is finitely presented. I was wondering whether this holds in full generality for algebraic theories, and if not, which parts ...
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1answer
51 views

Dual, Projective Modules, and Isomorphisms

Given a finitely generated right $R$-module $M$ with a generating set $\{m_i\}$, can one define a map $M$ to its dual left-module of right module functions by $m_i \mapsto m_i^*$ in analogy with the ...
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29 views

Weakest condition on module such that free prime localizations imply projective?

I know that for a finitely presented commutative module $M$, if all prime localizations $M_\mathfrak{p}$ are free, then $M$ is projective. Is there a weakening or a converse of this statement? What ...
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1answer
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If $M$ and $N$ are simple $R$-modules, and there is nonzero map from projective cover $P_N\to M$, is $N\cong M$?

Suppose $M$ and $N$ are simple modules over a commutative ring $R$, and $P_N$ is the projective cover of $N$. If there is a nonzero morphism $P_N\to M$, does this imply $N\cong M$? I'm not too ...
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Different definition of $K_0(R)$. Prove equivalence

I need to prove that the following two definitions of the zero-th $K$-theory group of a ring $R$ (with unit) are equivalent. def 1 the abelian group generated by the isomorphism classes $[P]$ of $...
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Help/Verification of Ext group calculation

I have been looking at extensions of irreducible representation over fields of positive characteristic. Specifically at the moment, to get the hang of things, I'm looking at $S_3$ over $\mathbb F_3$, ...
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1answer
73 views

If $P$ is projective and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$

I need to prove that if $P$ is a projective module over $\mathbb Z$ and $A,B$ are direct summands of $P$ then $A\cap B$ is a direct summand of $P$. This in turn would imply if $A$ and $B$ are ...
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1answer
47 views

Proving $\Bbb Z_p$ has no projective cover.

I need to prove that $\mathbb Z_p$ has no projective cover when seen as a module over $\mathbb Z$. I would appreciate your help. A projective cover of $M$ is a projective module $P$ along with a ...
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19 views

Two quotients of projective modules are equal, prove the crossed direct sums of the projective modules and kernels are isomorphic. (Schanuel's Lemma)

We have the following: $$\alpha_1:P_1\rightarrow M$$ $$\alpha_2:P_2\rightarrow M$$ with $\alpha_1,\alpha_2$ surjective, and $P_1,P_2$ projective. Prove $$P_1\oplus \ker(\alpha_2) \cong P_2\oplus \...