Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 24 / 3 / b

The intended statement uses a nontrivial Non-Archimedean absolute value. Under that convention choose $t$ with $0<|t|<1$. A compact neighborhood $C$ of zero contains an open ball of some radius $r>0$. For sufficiently large $N$, $t^N{\mathcal{O}}_K$ is contained in that ball. The valuation ring is closed, so $t^N{\mathcal{O}}_K$ is a closed subset of $C$ and is compact. Scaling shows that ${\mathcal{O}}_K$ itself is compact.

Its maximal ideal $\mathfrak{m}=\{x:|x|<1\}$ is open. The residue field $k={\mathcal{O}}_K/\mathfrak{m}$ is therefore discrete and compact, hence finite: the open cover by its singleton sets has a finite subcover. Since $\mathfrak{m}$ has finite index, its complement is a finite union of open cosets. Thus $\mathfrak{m}$ is also closed and compact.

The continuous absolute value on a field attains its maximum on $\mathfrak{m}$. Nontriviality supplies a nonzero element there, so this maximum is $\rho =|\pi |$ for some $\pi \ne 0$, with $0<\rho <1$. If $x\in \mathfrak{m}$, then $|x/\pi |\le 1$, so $\mathfrak{m}=\pi {\mathcal{O}}_K$.

For arbitrary $x\ne 0$, choose the integer $n$ for which ${\rho }^{n+1}<|x|\le {\rho }^n$. Then $\rho <|x/{\pi }^n|\le 1$. This element cannot lie in $\mathfrak{m}$, by maximality of $\rho$, so its absolute value on a field is one. Therefore $|x|={\rho }^n$. This proves discrete valuation from nontrivial local compactness:

The nontrivial valuation in the local compactness criterion is essential. If the trivial absolute value is admitted, the displayed source claim is false: take $\mathbb{Q}$ with that absolute value on a field. It is complete and discrete, hence locally compact, but its valuation ring and residue field are both $\mathbb{Q}$, which is infinite. The intended nontrivial case has been proved above.