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I am reading about modules and the author (my professor) asserts that the only finite free $\mathbb{Z}$-module is the trivial one. Knowing that abelian groups correspond to $\mathbb{Z}$-modules, we conclude that there are lots of finite modules which aren't free.

However, there's something stuck on my head that I can't falsify. Take a finite cyclic abelian group, which we know it is isomorphic to $\mathbb{Z}_{n}$. These abelian groups are generated by $1_{n}$. So, why aren't these finite cyclic groups a free module?

aadcg
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  • It is a basis. Having a basis is not the only condition for being a free module. Refresh yourself on the definition of "free module." – Thomas Andrews Jun 12 '17 at 16:07
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    As a Z-basis? But $n\cdot 1_n=0$. – Wuestenfux Jun 12 '17 at 16:08
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    @ThomasAndrews: The definition I have says that a free module admits a basis... What's missing? – aadcg Jun 12 '17 at 16:12
  • The specific proof depends on how you've defined "free," although all the definitions are equivalent in the end. – Thomas Andrews Jun 12 '17 at 16:12
  • Everything in the "..." is what is missing. Given the entire definition. – Thomas Andrews Jun 12 '17 at 16:12
  • @Wuestenfux True. But $n\in\mathbb{Z}$, so it not fixed whereas the $n$ featuring in $1_{n}$ is. – aadcg Jun 12 '17 at 16:15
  • I solved my problem. I don't understand why you didn't immediately said that my reasoning is wrong due to the fact that this isn't a basis - it fails to be linearly independent. – aadcg Jun 12 '17 at 16:22
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    @AAlexandre I don't know what your definition of basis is, but here is a good one. A collection of elements $m_i\in M$ is a basis for an $R$-module $M$ if $\sum r_i m_i=0$ implies $r_i =0$ for all $i$ . The fact that $n\cdot 1_n=0$ but $n\neq 0$ as an element of $\mathbb Z$ means that this is not a basis over $\mathbb Z$ (although it will be a basis over $\mathbb Z/n$. That's the thing, something can be free over one ring but not free over another.) – Aaron Jun 12 '17 at 16:32
  • That's a good definition of linear independence – D_S Jun 13 '17 at 12:29

1 Answers1

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A finite abelian group $G$ is not free as a $\mathbb{Z}$-module.

Indeed, suppose $|G|=n$ and let $p$ be a prime not dividing $n$.

Take an abelian group $H$ having an element $x$ of order $p$. Then there is no homomorphism $f\colon G\to H$ such that $x$ belongs to the image of $f$, because such image is annihilated by $n$, whereas $x$ is not.

More generally, a finitely generated free module over the ring $R$ has the form $R^n$, for some $n$.

egreg
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