Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 2

Suppose first that the valuation is discrete and the residue field $k$ is finite. For a uniformizer $\pi$, every quotient ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$ is finite, and completeness givesThis inverse limit is compact. Since ${\mathcal{O}}_K$ is a compact neighborhood of zero, $K$ is locally compact.

Conversely, local compactness gives a compact ball about zero, which can be rescaled to make ${\mathcal{O}}_K$ compact. Its distinct residue classes are disjoint open balls of radius below one, so compactness forces the residue field to be finite. Cover ${\mathcal{O}}_K$ by finitely many balls of some radius $r<1$. Applying the ultrametric inequality to centers lying in the maximal ideal produces $\rho <1$ such that every nonunit has absolute value at most $\rho$. Hence the value group has a largest value below one, and the valuation is discrete. This proves the local compactness criterion for a complete non-Archimedean field.

An algebraically closed valued field has an $n$th root of every element. If its valuation were discrete and $\pi$ were a uniformizer, then $v(\sqrt[n]{\pi })=1/n$ would contradict discreteness. Therefore an algebraically closed non-Archimedean field cannot be locally compact.

If $|\cdot |$ is non-Archimedean, then $|n|=|1+\cdots +1|\le 1$. Conversely suppose $|n|\le M$ for every integer $n$. The binomial theorem and the ordinary triangle inequality giveTaking $N$th roots and letting $N\to \infty$ proves $|x+y|\le \max (|x|,|y|)$. This is the bounded-integer criterion for a non-Archimedean absolute value.

In characteristic $p$, the image of $\mathbb{Z}$ is the finite prime field ${\mathbb{F}}_p$, so every absolute value is bounded on it. Thus every absolute value on ${\mathbb{F}}_p(t)$ is non-Archimedean.

The polynomialhas no root in ${\mathbb{F}}_3$ and is therefore irreducible. Definefor any fixed $c>1$. Its valuation ring has residue fieldThe completion at an irreducible polynomial over a finite field identifies the completion with ${\mathbb{F}}_{27}((T))$, where $T$ corresponds to the uniformizer $h(t)$.