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

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?

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.
• Wow, things become clear. Thank you. Neukirch's book p.112 is associated reference. Can I ask more? Why can we view any natural number $n$ as an element of $\mathbb{Z}_p$? I think that it is elementary question. Commented Oct 24, 2023 at 11:10
• I guess that, upto embedding $\mathbb{Q} \to \mathbb{Q}_p$, $a \mapsto (a,a,a, \dots)$ ( c.f. Neukirch's book p.110), $n$ can be viewed as a (nonzero) element of $\mathbb{Q}_p$ ( $(n,n, \dots )$ is not null-sequence - c.f. Neukirch's book p.110 -? ) and we can apply your argument ? True? Commented Oct 24, 2023 at 11:30
• Yes, that's exactly the canonical embedding. Since $\nu_p$ on $\Bbb{Q}_p$ is really an extension of the usual $p$-adic valuation $\nu_p$ on $\Bbb{Q}$, meaning they agree on (the canonical embedding of) $\Bbb{Q}$, and $\nu_p$ on $\Bbb{Q}$ is really just the power of $p$ in the factorisation, it becomes clear that the image of $\Bbb{Z}$ lands in $\Bbb{Z}_p$ under the embedding. Commented Oct 24, 2023 at 12:44