$\# (\mathbb{Z}_p/n\mathbb{Z}_p)=p^{v_p(n)}$, where $v_p$ is the p-adic exponential valuation?

Question

Let $\mathbb{Q}_p$ be the field of $p$-adic numbers with $\mathbb{Z}_p = \{ x\in \mathbb{Q}_p : |x|_p \le 1\} $ its valuation ring ( c.f. Neukirch, Algebraic Number Theory, p.111, (2.3) Proposition ) (or can be regared as the set of $p$-adic integers (?)). Let $v_p : \mathbb{Q} \to \mathbb{Z} \cup \{ \infty\}$ be the $p$-adic exponential valuation ( c.f. Neukirch, p.107 ) or its extension to $\mathbb{Q}_p \to \mathbb{Z} \cup \{ \infty\}$ ( c.f. Neukirch, p.110 ~ 111 ).

Let $n$ be a natural number.

Then, my question is,

$\# (\mathbb{Z}_p/n\mathbb{Z}_p)=p^{v_p(n)}$ ? True? If so, why?

This question originates from following linked question that I Proposed : Understanding the Neukirch, Algebraic Number Theory, p.142, (5.8) Corollary.. There I asked why the second equality in $(3)$ is true.

Can anyone help?

Answer 1

This is true, in fact true for any $n\in \Bbb{Z}_p$. Numbers in $\Bbb{Q}_p$ have the fantastic property that they can always be written as $p^k\varepsilon$, where $k\in\Bbb{Z}$ and $\varepsilon$ is a $p$-adic unit, i.e. it has an inverse in $\Bbb{Z}_p$. If we restrict to $p$-adic integers, then $k\geq 0$. Of course, this $k$ is exactly the $p$-adic valuation of the number. So if $n=p^k\varepsilon$, then $\nu_p(n)=k$ and $$n\Bbb{Z}_p = p^k\varepsilon\Bbb{Z}_p = p^k\Bbb{Z}_p.$$ Now, it's a standard result that $\Bbb{Z}_p/p\Bbb{Z}_p\cong \Bbb{Z}/p\Bbb{Z}$, perhaps, you've seen this already. One basically shows that the composite $\Bbb{Z}\hookrightarrow\Bbb{Z}_p \twoheadrightarrow \Bbb{Z}_p/p\Bbb{Z}_p$ is surjective, and check that the kernel is what we suspect. But all this generalizes very neatly to any power of $p$, giving you that $\Bbb{Z}/p^k\Bbb{Z}\cong \Bbb{Z}_p/p^k\Bbb{Z}_p$, and then the cardinality result follows immediately.