# n-th power residues in p-adic fields

How to see that number of n-th power residue classes of a p-adic field K (elements of $K^*/K^{*n}$) is finite? I know how to prove using Hensel lemma that all p-adic integers sufficiently close to 1 are n-powers but cannot apply it to prove the statement.

Let $\pi$ be a prime element, i.e. generator of the maximal ideal of the ring $\mathfrak o_K$ of integers of $K$. Then $K^*=\pi^{\Bbb Z}\oplus\mathfrak o_K^*$, where $\mathfrak o_K^*$ is the group of units, which is compact. But $z\mapsto z^n$ is an open mapping, you’ve already proved this, and so the image is open in the compact group $\mathfrak o_K^*$, consequently of finite index.
• Can I ask why $z\mapsto z^n$ is an open map? – Dr. Heinz Doofenshmirtz 8 hours ago
• Because we’re dealing with a topological group, it’s sufficient to decide this question in a neighborhood of the identity element $1$. But the neighborhood, describable as $\{z: v(z-1)>M\}$ for some $M$, gets mapped onto $\{z: v(z-1)>M'\}$, where $M'=M+v_p(n)$. You’ll need $M\ge2$ for this, I think. – Lubin 8 hours ago
The multiplicative group structure of $K$ can be written down explicitly (see e.g. Neukirch - Algebraic Number Theory - 2.5.7) $$K^*\cong\mathbb{Z}\times\mathbb{Z}/(q-1)\times\mathbb{Z}/p^a\times(\mathbb{Z}_p)^d,$$ where $q$ is the size of the residue field, $p$ the characteristic of the residue field of $K$, $d$ the extension degree to $\mathbb{Q}_p$, and $a$ a non-negative integer.
So the finitness of $K^*/K^{*n}$ reduces to the finitness of $(\mathbb{Z}\times(\mathbb{Z}_p)^d)/(n\mathbb{Z}\times (n\mathbb{Z}_p)^d)$, and hence the finitness of $\mathbb{Z}_p/n\mathbb{Z}_p$. By writting $n=p^bn'$, $n'$ invertible in $\mathbb{Z}_p$, we see $\mathbb{Z}_p/n\mathbb{Z}_p=\mathbb{Z}_p/p^b\cong(\mathbb{F}_p)^b$.