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Let $\mathbb{Z}_p$ denote the p-adic intergers.I know this is torsion-free Abelian group(i.e. It is flat $\mathbb{Z}$ module).

  • I think it is not free Abelian group,but I don't know how to prove this.?

  • Is $Ext^1_{\mathbb Z}(\mathbb{Z}_p,\mathbb {Z})\neq 0$？

  • First I select a s.e.s $0\rightarrow \mathbb{Z}_p\rightarrow \prod\mathbb{Z}/p^i\mathbb{Z}\rightarrow^{1-shift}\prod\mathbb{Z} /p^i \mathbb{Z} \rightarrow 0$. But $Ext^1_{\mathbb Z}(\prod \mathbb {Z},\mathbb{Z}\neq 0$,this can see my top vote.Then how should I do?

I am confused.I need your help.Thanks!

Jian
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    Note that $\mathbb Z_p$ is $q$-divisible for any prime $q \neq p$. A free abelian group is certainly not $q$-divisible for any prime $q$. – MooS Sep 04 '17 at 09:57
  • @MooS nice observation！thank you. – Jian Sep 04 '17 at 10:11
  • @MooS Is the second question true? – Jian Sep 04 '17 at 10:14
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    For the second question, consider the exact sequence $0\to \mathbb{Z}p\stackrel{p}{\to}\mathbb{Z}_p\to\mathbb{Z}/p\mathbb{Z}\to 0$ and observe that $Hom{\mathbb{Z}}(\mathbb{Z}p,\mathbb{Z})=0$ and $Ext^1{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z},\mathbb{Z})\neq 0$ to show that your ext is not zero. – Mohan Sep 04 '17 at 13:09
  • @Mohan thank you very much.the s.e.s you give is very perfect! – Jian Sep 04 '17 at 15:16
  • Related: https://math.stackexchange.com/q/2399486/96384 – Torsten Schoeneberg Dec 01 '22 at 20:02

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