6
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Schepler Apr 19 '19 at 22:15
  • 1
    $\begingroup$ Not quite the same question but very similar: math.stackexchange.com/questions/2664460/… $\endgroup$ – Eric Wofsey Apr 19 '19 at 22:35
  • $\begingroup$ Thanks @EricWofsey , will look into that. $\endgroup$ – Leo Apr 19 '19 at 22:38
  • $\begingroup$ @DanielSchepler See the link in Eric Wofsey's comment $\endgroup$ – Hagen von Eitzen Apr 19 '19 at 22:51
  • $\begingroup$ Related: mathoverflow.net/questions/48899/… $\endgroup$ – Arturo Magidin Apr 20 '19 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.