If $|\cdot|$ is non-Archimedean, then $|n|=|1+\cdots+1|\leq1$. Conversely suppose $|n|\leq M$ for every integer $n$. The binomial theorem and the ordinary triangle inequality give
$$
|x+y|^N\leq (N+1)M\max(|x|,|y|)^N.
$$
Taking $N$th roots and letting $N\to\infty$ proves $|x+y|\leq\max(|x|,|y|)$. 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.

Solution

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

An absolute value on a field is non-Archimedean exactly when its values on the image of $\mathbb Z$ are bounded. The reverse implication follows by applying the ordinary triangle inequality to the binomial expansion of $(x+y)^n$ and taking $n$th roots as $n\to\infty$.


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