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

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 $h(t)$.