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Questions tagged [free-modules]

Use this tag for questions about free modules and related notions as projective modules or free abelian groups. This tag should be used together with the tags of abstract algebra and modules.

In abstract algebra, a free module is a module that has a basis, that is, a generating set consisting of linearly independent elements. Every vector space is a free module but, if the ring of the coefficients is not a division ring (not a field in the commutative case), then there exist non-free modules. On the other hand, free abelian groups are precisely the free modules over the ring $\Bbb Z$.

543 questions
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How can $\mathbb{R}/\mathbb{Z}$ be a free $\mathbb{R}$-module?

I'm a little confused here. On the one hand, $\mathbb{R}/\mathbb{Z}$ must be free as an $\mathbb{R}$-module, since any module over a field is free (being a vector space it has a basis). One the other hand, it can't be free, as it has torsion - e.g.…
Kimarokko
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Surjective homomorphism between a R-module and a free R-module

I'm trying to show that if $F$ is a free $R-$module and $g:M\rightarrow F$ is a surjective $R-$homomorphism then $M \simeq Ker(f)\oplus F$. What I have done so far: Let $X=\{x_i\}_{i\in I}$ be a base for $F$, define the $R-$homomorphism…
Vinícius Riter
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Module Theory, Free Modules, Finitely Generated Modules

I want to prove that if R is a PID and M is a finitely generated R-module, then there exist a free R- module F such that $M \cong F \oplus M_{\text{tors}}$ So, in order to prove the above I set $M/M_{\text{tors}} = F$ and since R is a PID then it…
dota
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Every projective module is a submodule of a free module?

I've seen this statement on the internet but I could not find a proof. Actually this is true for any module I think. Can a proof be given as follows? Let $M$ be an $R$-module. Take a generating set $X$ of $M$ over $R$. Then consider the free…
Philip Johnson
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When $R$-Module of Rank $r$ has a submodule isomorphic to $R^r$?

Let $R$ be an integral domain,$M$ is a finitely generated $R$ module. Prove that rank$(M)=r$ iff $M$ has a free submodule $N \equiv R^r$ , such that $M/N$ torsion . If $R$ is a PID then $N$ may be chosen so that $0\to N\to M\to M/N\to 0$ splits. If…
Infinity
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Extending a morphism among R-modules using a basis to make a diagram commute

let M,N R-modules, and L a free R-module. consider the morphism g:L⟶N,f:M⟶N, f epimorphism. Show that exist h:L⟶M such that f∘h=g I try to define h in some basis, with that i can use a theorem that extend this morphism in L, but since f isn't…
Luiz Guilherme De Carvalho Lop
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why we need "torsion-free"

This is from P 134 of Rotman's Homological Algebra book. If R is a PID, then every torsion-free R-module is flat. Proof. If R is a PID, then every finitely generated R-module M is a direct sum of cyclic modules. If M is torsion-free, then it is a…
scsnm
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how does P have n generators?

This is from Page 101 of Rotman's book on homological algebra. I am not sure for the proof of ii, why there are as many generators of P as basis elements? Any help would be appreciated! Thank you!
scsnm
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Passing from $\mathbb{F}_p$ to $\mathbb{Z}$.

I am reading Serre's book on Lie Algebras and Lie Groups. On Lemma 4.3, he states that If $E$ is a finitely generated $\mathbb{Z}$-module and $\dim(E \otimes_{\mathbb{Z}} \mathbb{F}_p)$ over $\mathbb{F} = \mathbb{Z} / p\mathbb{Z}$ is…
André Caldas
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The module $Ra$ is free on $A$?

I was recently reading this question about free modules. In this post, the definition of a free module is given as An $R$-module $F$ is said to be free on the set $A \subset F$ if for every nonzero $x \in F$, there exist unique $r_1, r_2, \cdots,…
Sam Y.
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