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I was reading this article http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/ostrowskiF(T).pdf in which they prove the Ostrowsky theorem for $K(x)$ where K is a field.

However I noticed that they assume that the absolute value in $K(x)$ is non-archimedean and I don't really find an easy way to prove this.

McNuggets666
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    They assume that the valuation is trivial on $K$. Isn't it a fact that if a valuation is bounded on the prime field then it is non-archimedean? – Jyrki Lahtonen Dec 03 '17 at 21:28
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    @JyrkiLahtonen: I must have upvoted your comment years ago, but only now it dawned on me that it (is true but) should be strengthened: A valuation on a field $F$ is non-archimedean/ultrametric if and only if it is bounded on the image of $\mathbb Z$ in $F$ under the canonical ring homomorphism $\mathbb Z \rightarrow F$. For positive characteristic, of course that is the prime field, but in char. $0$, your comment takes care only of the trivial value on $\mathbb Q$, whereas now we have subsumed all the $p$-adic ones as well. Cf. second link in my answer below. – Torsten Schoeneberg Apr 15 '21 at 03:47
  • Thanks for the extras @TorstenSchoeneberg. I seem to have upvoted your answer back in the day :-) – Jyrki Lahtonen Apr 15 '21 at 06:48

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To expand on Jyrki Lahtonen's comment:

On the first page of the paper you link to, "example 1" actually shows that this is not true in general. E.g. on $\mathbb{Q}(x)$ one can define lots of archimedean valuations. Namely, choose any number $\alpha \in \mathbb{C}$ which is transcendent (over $\mathbb{Q}$), and check that sending $x$ to $\alpha$ defines an isomorphism $\mathbb{Q}(x) \simeq \mathbb{Q}(\alpha) \subset \mathbb{C}$. Then restrict the usual absolute value of $\mathbb{C}$ to that subfield and pull it back to $\mathbb{Q}(x)$. Compare also the answer ("Oh yes there are!") to the related (but ill-posed) question: There are no archimedean function fields.

That is why after the examples, that paper explicitly works with the extra assumption that the restriction of the valuation to $K$ is trivial, meaning that $|a| = 1$ for all $x \in K^*$. Under this assumption, the valuation on $K(x)$ is indeed automatically non-archimedean. Depending on what your definition of "non-archimedean" is, this is more or less trivial, as it means that for all $a\in K$,

$|na| = 1$ for all $n\in\mathbb{N}$.

If your definition of "non-archimedean" is not already checked by this fact, but rather makes use of the strong triangle inequality / ultrametric, one can see e.g. here that all those are already implied by the above fact.

Torsten Schoeneberg
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