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Let $p$ be a prime. Is $\mathbb{Z}_p$ a free $\mathbb{Z}_{(p)}$ module?

Here are some more details for those who are interested:

Here ${\mathbb Z}_p$ are the ${\mathbb p}$-adic integers (so the completion of ${\mathbb Z}$ with respect to the $p$-adic topology).

The ring ${\mathbb Z}_{(p)}$ is the localization of ${\mathbb Z}$ with respect to the prime ideal $(p)$. This is a discrete valuation ring.

Note that any integer $m$ different from $p$ is invertible in ${\mathbb Z}_p$, so we have a morphism ${\mathbb Z}_{(p)} \to {\mathbb Z}_p$. In particular the latter is a ${\mathbb Z}_{(p)}$-module.

For modules of finite type over ${\mathbb Z}_{(p)}$ one can check very quickly that they are free, as soon as they are torsion free. For modules that are not of finite type, one cannot check this as fast, because it is false in general (e.g. ${\mathbb Q}$ is not of finite type and not free, yet it is torsion free over ${\mathbb Z}_{(p)}$).

Hence my question: Is ${\mathbb Z}_p$ free as a ${\mathbb Z}_{(p)}$ module?

(Note that am I not asking if it is free of finite rank, as that would be clearly false).

Arturo Magidin
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mnr
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1 Answers1

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No idea why some people voted to delete my answer, without even leaving a comment.

If $\Bbb{Z}_p$ is a free $\Bbb{Z}_{(p)}=(\Bbb{Z}-(p))^{-1}\Bbb{Z}$ module of rank $n$ (it can be $n=\infty$) then the quotient group $$\Bbb{Z}_p/p\Bbb{Z}_p\cong \Bbb{Z}_{(p)}{}^n/p\Bbb{Z}_{(p)}{}^n\cong \Bbb{F}_p{}^n$$ so that $n=1$ which is absurd.

Whence $\Bbb{Z}_p$ is a non-free $\Bbb{Z}_{(p)}$-module.

I don't think there is any easier argument. If I am missing something then there is the comment box..

reuns
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  • Why $\mathbb{Z}p$ can't be free $\mathbb{Z}{(p)}$-module of rank $1$ ? Is it because $p$ generates $\mathbb{Z}p$ while $p \notin \mathbb{Z}{(p)}$ ? – MAS Mar 08 '21 at 16:54
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    Well, @Why , the two aren’t even of the same cardinality. – Lubin Mar 08 '21 at 17:06
  • @Why $\Bbb{Z}p$ are the $p$-adic integers, topologically generated by $1$ (ie. the closure of $\Bbb{Z}.1$) or any $a\ne 0\bmod p$, and $\Bbb{Z}{(p)}$ are the rational numbers with denominator not divisible by $p$. It is well-known that $\Bbb{Z}_p$ contains many elements not in $\Bbb{Q}$ like $\sqrt{1+p^2}$ (Hensel lemma). – reuns Mar 08 '21 at 17:40