# Does every non-archimedean absolute value on field take value in $\mathbb{Q}$

Let $K$ be a field, a non-archimedean absolute value is defined to be a map $K\to \mathbb{R}$ satisfying $|x|=0\Rightarrow x=0$, $|x|\cdot|y|=|xy|$ and $|x+y|\leq\max(|x|,|y|)$.

Is there an example that the valuation of a non-archimedean field not take value in $a^\mathbb{Q}$ for any fixed real number $a$?

• Writing \max rather than \mathrm{max} results in proper spacing in things like $4\max S$ and in $4\max(S)$ (in the latter there's less space to the right of $\max$), and affects the format of the subscript in things like $\displaystyle \max_{x\in S} f(x)$. (That is how it appears in a displayed setting; in an inline setting it appears as $\max_{x\in S} f(x)$, with the subscript on the right rather than just below $\max$.) \max is standard usage and I edited accordingly. $\qquad$ Feb 1, 2016 at 18:07
• Your question is not clear: are you asking whether we can generalize to irrational numbers or to some other group outside the real numbers? Feb 1, 2016 at 18:08
• @crostul Thanks for clarification! I hope the group lie in real numbers but not in rational numbers
– user93417
Feb 1, 2016 at 18:59
• I think you want your map to go to $\mathbb{R}_{\ge 0}$, and you want $K^*$ to (not) take value in $a^\mathbb{Q}$. Dec 24, 2017 at 4:59

Think to $$x^{\sqrt{2}}$$ as a function $$(0,\infty)\to \Bbb{R}$$, no problem to look at the ring $$\Bbb{R}[x,x^{\sqrt2}]$$ and its fraction field $$K$$ with the natural valuation $$v(\sum_{n,m}c_{n,m}x^{n+m\sqrt2})=\inf \{ n+m\sqrt2, c_{n,m}\ne 0\}, \qquad |f/g|=2^{v(f)-v(g)}$$

Yes (to the question in your post, not in the title). Here is an example: Choose your favourite prime $p$, your favourite positive number $r\notin p^\mathbb{Q}$, and define the following absolute value on the polynomial ring $\mathbb{Q}[t]$:

$$|\sum_{i=0}^n a_i t^i| := \max (|a_i|_p r^i)$$

where $|\cdot|_p$ is the usual $p$-adic absolute value on $\mathbb{Q}$. Extend it multiplicatively to the field of fractions $K:=\mathbb{Q}(t)$. We have

$$|K| = \{0\} \cup p^\mathbb{Z}\cdot r^\mathbb{Z}.$$

In particular both $p$ and $r$ are in $|K^*|$, but if there were an $a\in \mathbb{R}$ with $p \in a^\mathbb{Q} \ni r$, it would follow that $r \in p^\mathbb{Q}$, contradiction.

Note that of course there are other non-archimedean (and non-trivial) absolute values on that field such that the value group $|K^*|$ is contained in $a^\mathbb{Q}$. A follow-up question would be: Are there any fields $K$ such that every non-trivial nonarchimedean value on $K$ has a value group that is not contained in some $a^\mathbb{Q}$?

• For the last point: treat the algebraic extensions of $\Bbb{Q,F_p}$ separately, (with the axiom of choice giving a transcendental basis) any other field $K$ is an algebraic extension of $k(x)$ for some subfield $k$ and $x\in K$ so the discrete valuation on $k[x]_{(x)}$ extends to a $\Bbb{Q}$-valued valuation on $K$. Mar 7, 2021 at 16:17

Hint: Any power of a non-archimedian absolute value is again an absolute value. Power by a real number. See if you can arrange irrational values that way.

• Thank for the example! But this is not what I wanted, so I changed my question to exclude the possibility..
– user93417
Feb 1, 2016 at 19:13

Is there an example of a non-Archimedean valuation that does not take values in $a^\mathbb{Q}$ for any fixed real number $a$?

The answer is no. In fact, if $|\cdot|$ is the trivial valuation, then it takes values in $a^\mathbb{Q}$ for $a=0$ and $a=1$. If $|\cdot|$ is not the trivial valuation, then there exists $x\in K$ such that $|x|\not\in\{0,1\}$. Thus, $|x|^\mathbb{Z}\subset|K|$.

But maybe the question that Qixiao had in mind was: Is there an example of a non-Archimedean valuation over a field $K$ such that $|K|\subset a^\mathbb{Q}$ for no real number $a$? I do not know the answer to this second question.

• See my answer ${}{}$ Mar 7, 2021 at 16:05