Why can we write $\mathbb{Z}_p$ as the disjoint union of balls $a+p\mathbb{Z}_p$

Question

In this chapter, Example 2.1 notes that we can write $\mathbb{Z}_p$ as $$p\mathbb{Z}_p\cup(1+p\mathbb{Z}_p)\cup\cdots\cup (p-1+p\mathbb{Z}_p).$$ I understand that $\mathbb{Z}_p=\{x\in\mathbb{Q}_p:|x|_p\leq 1\}$ and that $p\mathbb{Z}_p=\{x\in\mathbb{Q}_p : |x|_p<1\}=\{x\in\mathbb{Q}_p : |x|_p\leq p^{-1}\}$, but I'm having trouble seeing why $\mathbb{Z}_p\subset \{a+p\mathbb{Z}_p\}$ for $a$ ranging between $0$ and $p-1$. I understand that each of these is the closed ball of radius $p^{-1}$ centered at $a$, but I don't see how that leads to these balls being a cover for $\mathbb{Z}_p$. It must be the case that any point in $\mathbb{Z}_p$ can be written as $a+x$ for $a$ between $0$ and $p-1$ and $|x|_p\leq p^{-1}$, but I don't see why that's true. While showing this mathematically is helpful, I'm trying to understand it from a conceptual point of view.

Answer 1

You can think of $\Bbb{Q}_p$ as numbers written in base $p$ notation that extend infinitely to the left of the $p$-ary point, but finitely to the right. So you'd think of the example: $$ x = a_mp^m + a_{m+1}p^{m+1} + \ldots $$ in your linked chapter as: $$ x = \ldots a_3a_2a_1a_0.a_{-1} \ldots a_{m} $$ Thinking of it this way, the arithmetic operations are done much as one does for finite base $p$ numbers, but carrying on to the left indefinitely. In particular, multiplication by $p$ is a left shift. $\Bbb{Z}_p$ comprises the $x$ as above where $m \ge 0$, i.e., where there are no digits after the point. But then $$ x = \ldots a_3a_2a_1a_0.0 = a_0 + p(\ldots a_3a_2a_1a_1.0) \in a_0 + p\Bbb{Z}_p $$

[Postscript: in case anyone is worried about how division works: you work from right to left killing successive least-significant digits.]

Answer 2

I think that in this case, a very good way to think is that $p\Bbb Z_p$ is a subgroup of $\Bbb Z_p$, indeed of index $p$, since $\Bbb Z_p/p\Bbb Z_p\cong\Bbb F_p$, the field with $p$ elements. But the cosets of a subgroup are pairwise disjoint, so all you need to do is find a representative of the $p$ cosets. The choice is up to you, but there are natural candidates, the ones you know.

(I always recommend the way of looking at $p$-adic numbers in the way that @Rob has in his answer, so I certainly do not deprecate what he had to say.)