Let $p$ be a prime and let $\mathbb{Z}_p$ denote the $p$-adic integers. One has a canonical inclusion of rings $$\mathbb{Z}_{(p)}\longrightarrow\mathbb{Z}_p$$ given by identifying the rational integers in $\mathbb{Z}_p$.

What is the quotient group $\mathbb{Z}_p/\mathbb{Z}_{(p)}$?

It was suggested to me, without proof, that the quotient is a $\mathbb{Q}$-vector space (of uncountable rank). I have been trying to verify this in detail.

My approach was the following:

  1. Show that the quotient is $p$-divisible
  2. Show that the quotient is torsion-free

For then the quotient is torsion-free and divisible, and the result follows.

However, I can neither prove 1 nor 2; my knowledge of the $p$-adic integers is a bit weak. How might I prove these statements, or should I try another approach?

  • $\begingroup$ But $\Bbb Z[1/p]$ is $p$-divisible and torsion-free, but not a $\Bbb Q$-space. Did you mean $q$-divisible for every prime $q$? $\endgroup$
    – Lubin
    Oct 22 '15 at 4:02
  • $\begingroup$ Well the $p$-adics are already $q$-divisible for every prime $q\not=p$, and this property is inherited by quotients. $\endgroup$
    – H.Durham
    Oct 22 '15 at 7:20

The approach works, and ends up being elementary.

Firstly: for an abelian group $A$, one has $A\otimes \mathbb{Z}/p\mathbb{Z}=A/pA$. Thus $p$-divisibility is equivalent to the tensor product with $\mathbb{Z}/p\mathbb{Z}$ vanishing.

In our case we have $$\frac{\mathbb{Z}_p}{\mathbb{Z}_{(p)}}\otimes \frac{\mathbb{Z}}{p\mathbb{Z}}\cong \frac{\mathbb{Z}_p}{\mathbb{Z}_{(p)}+p\mathbb{Z}_p},$$ but this latter quotient is $0$, since any $\gamma\in\mathbb{Z}_p$ may be written as

$$\gamma=a_0+p\sum_{i\geq1}a_ip^{i-1}\in \mathbb{Z}_{(p)}+p\mathbb{Z}_p.$$

To see why the quotient is torsion-free, suppose that $\gamma\in\mathbb{Z}_p$ has $n\gamma\in\mathbb{Z}_{(p)}$. If $p\!\!\not|\;n$ then $\frac1n$ exists and $\gamma\in\mathbb{Z}_{(p)}$. Thus we may reduce to the case that $n=p$. Write $$p\gamma=\frac km$$ for $p\!\!\not|\;m$. If now $p$ divides $k$, we are done. If not, we have that $p\gamma$ is a unit, which is impossible.


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