# Quotient of $\mathbb{Z}_p$ by the rational integers

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?

• 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 $q$? Oct 22 '15 at 4:02
• Well the $p$-adics are already $q$-divisible for every prime $q\not=p$, and this property is inherited by quotients. Oct 22 '15 at 7:20

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.