$R^{\mathbb N}$ as a free $R$-module.

Asked Apr 19 '19 at 21:47

Active Apr 19 '19 at 22:53

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Suppose that $R$ is a commutative ring. I'm wondering if the space $R^{\mathbb N}$ is a free $R$ module.

I know how to prove that it is not a free $R$ module in the case of $R = \mathbb Z$. But the proof I know/could think of, uses facts specific of $\mathbb Z$ such as it being an Euclidean ring. So I was unable to recycle this idea for a generic $R$. Also I know that if $R$ is a field, then every $R$-module is free. So in particular $R^{\Bbb N}$ is a free $R$ module. So the question is,

What happens when $R$ is a commutative ring but not a division ring?

As an example $R=\Bbb Z_n$ if it is too broad a question.

edited Apr 19 '19 at 22:53

asked Apr 19 '19 at 21:47

leolerena

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I've heard that $\mathbb{Z}^{\mathbb{N}}$ is not a free abelian group (despite being torsion free) though personally I haven't seen the proof. – Daniel Schepler Apr 19 '19 at 22:15
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Not quite the same question but very similar: https://math.stackexchange.com/questions/2664460/when-is-every-direct-product-of-a-ring-also-a-free-module – Eric Wofsey Apr 19 '19 at 22:35
Thanks @EricWofsey , will look into that. – leolerena Apr 19 '19 at 22:38
@DanielSchepler See the link in Eric Wofsey's comment – Hagen von Eitzen Apr 19 '19 at 22:51
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Related: https://mathoverflow.net/questions/48899/can-the-i-fold-direct-product-be-free – Arturo Magidin Apr 20 '19 at 02:32

$R^{\mathbb N}$ as a free $R$-module.

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