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

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.

• 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
• Not quite the same question but very similar: math.stackexchange.com/questions/2664460/… – Eric Wofsey Apr 19 '19 at 22:35
• Thanks @EricWofsey , will look into that. – Leo Apr 19 '19 at 22:38
• @DanielSchepler See the link in Eric Wofsey's comment – Hagen von Eitzen Apr 19 '19 at 22:51
• – Arturo Magidin Apr 20 '19 at 2:32