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Two absolute values on a field are equivalent absolute values when they induce the same topology. For nontrivial absolute values on a field this is equivalent tofor a fixed real number $c>0$. The trivial absolute value is equivalent only to itself.

Restriction of a Non-Archimedean absolute value plainly preserves the ultrametric inequality. Conversely, if the restriction is non-Archimedean, every integer, interpreted in either field, has absolute value on a field at most one. In particular every binomial coefficient has absolute value on a field at most one. The triangle inequality applied to the binomial expansion gives, for $x,y\in L$ and every positive integer $N$,Raise both sides to the power $1/N$ and let $N\to \infty$. Since $(N+1{)}^{1/N}\to 1$, this yieldsThis is the bounded-integer criterion for a non-Archimedean absolute value. The proof works for arbitrary field extensions, not just algebraic ones, and in positive characteristic as well.

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.