Can $\Bbb{Z}$ be regarded as a $\Bbb {Z}_p$-module?

Question

Can $\Bbb{Z}$ be regarded as a $\Bbb {Z}_p$-module, where $\mathbb{Z}_p$ is the set of $p$-adic integers?

I know $\Bbb{Z}$ cannot be $\Bbb {Z}_p$-algebra, so I cannot make $\Bbb{Z}$ into $\Bbb {Z}_p$-module through natural way inherited from $\Bbb {Z}_p$-algebra structure.

Can I regad $\Bbb{Z}$ as $\Bbb {Z}_p$-module in some way?

Answer 1

No, it's not possible. If $q\in \mathbb{Z}$ is not divisible by $p$, then $1/q\in \mathbb{Z}_p$, so for any $\mathbb{Z}_p$-module $M$ and any $x\in M$, we must have $x = q\cdot (\frac{1}{q}x)$, so $x\in qM$, and $M$ is $q$-divisible as an abelian group.

But obviously $\mathbb{Z}$ is not $q$-divisible for any $q\neq \pm 1$.

Answer 2

To avoid ambiguity, I will call the p-adic integers $\mathbb{Z}_{pa}$.

Here's a simple question. What is the ideal $I\subset\mathbb{Z}_{pa}$ of scalars $x$ such that $x\cdot\mathbb{Z}=0$?

If $I=0$, then the elements $x\cdot 1$ are all distinct, so $\mathbb{Z}$ would be an uncountable set.

If $0<I<\mathbb{Z}_{pa}$, then $\mathbb{Z}$ would be a non-trivial module over $\frac{\mathbb{Z}}{p^n\mathbb{Z}}$, which is impossible.

The only possibility therefore is $I=\mathbb{Z}_{pa}$.