Questions tagged [projective-module]

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

We call a module $P$ over a ring $R$ projective if for every surjective $R$-module homomorphism $f : N \to M$, and every module homomorphism $g : P \to M$, there is a lift $h$. That is, there is a module homomorphism $h : P \to N$ with $fh = g$.

Alternatively, a module $P$ is projective if every short exact sequence $$0 \to A \to B \to P \to 0$$

of $R$-modules splits.

Projective modules can be viewed as generalizations of free modules, and every free module is projective.

Source: Projective module.

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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. I am stuck here. I saw some questions in SE, but…
lee
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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 $P \oplus N$ is free and finitely…
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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 Modern Algebra. Using theorems before exercises, because…
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Finitely generated projective modules are locally free

Let $A$ be a commutative noetherian ring, and let $M$ be a finitely generated projective $A$-module. It is well known and easy to prove that $M$ is locally free in the sense that for every $p \in\operatorname{Spec} A$, the module $M_p$ is a free…
the L
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Finitely generated projective module

Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free? What I know about a finitely generated projective module $M$ over a PID $R$ is isomorphic to $R^k\oplus R/(a_1)\oplus\dots\oplus…
Marso
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If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective

If over a commutative ring $R$ we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ are finitely generated projective? We have finite generation, because if $M\otimes N$ is generated by $\sum_i a_{ij} x_i^j\otimes y_i^j$, then…
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Projective module over a PID is free?

A common result is that finitely generated modules over a PID $R$ are projective iff they are free. Is the same true that an arbitrary projective module over a PID is free? I can't find this fact anywhere, so I suspect it is false, but I can't…
Hana Bailey
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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 to our question f.g. flat not free where all the…
Jack Schmidt
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Origin of the terminology projective module

Projective modules were introduced in 1956 by Cartan and Eilenberg in their book Homological Algebra. Does anyone know why they chose the word "projective"? Does it have something to do with the notion of projection?
PatrickR
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Lemma on infinitely generated projective modules

Is it true that every finitely generated submodule of a non-finitely generated projective over a (not necessarily commutative!) ring is contained in a proper summand? (Ideally there's a standard reference for this... :) )
Mariano Suárez-Álvarez
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Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.

Disclaimer: Though I have been re-reading my notes, and have scanned the relevant texts, my commutative algebra is quite rusty, so I may be overlooking something basic. I want to show $\mathbb{Q} \simeq \mathbb{Q}[x,y]/\langle x,y \rangle$ is not…
Derek Allums
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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. This implies that every short exact sequence $$0…
user58289
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In which algebraic theories do 'free' and 'projective' coincide?

Free models of algebraic theories are always projective objects in the category of models, but the converse is not always true. For instance, some (actually, all) projective modules are direct summands of free modules, and these need not be free. On…
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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, $\Bbb{Z}, \Bbb{Z}, \Bbb{Z}, \cdots$ whose product…
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Projective resolution of tensor product

Let $M,N$ are $R$-modules and $P^\bullet, Q^\bullet$ are their projective resolutions. Can we obtain a projective resolution of $M\otimes N$ using $P^\bullet, Q^\bullet$? If I understand correctly the homology groups of $P^\bullet\otimes Q^\bullet$…
user52045
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