# non-Archimedean Valued field extension of $\mathbb{R}$

Let $K$ be a field with non-Archimedean valuation $|\cdot|$. Suppose that $\mathbb{R}\subset K$.

Question 1: Is the restriction of $|\cdot|$ to $\mathbb{R}$ the trivial valuation?

I guess that the answer is yes, but I don't see an evident answer. When I try to organize my ideas I get other questions:

Definition: Two valuations $|\cdot|_1$ and $|\cdot|_2$ are dependent if there exists $\lambda>0$ such that $|\cdot|_1=|\cdot|_2^{\lambda}$.

According Ostrowski's theorem, if the restriction of $|\cdot|$ to $\mathbb{Q}$ is not trivial, then it is dependent on the p-adic valuation.

Question 2: How to prove that $|\cdot|\neq|\cdot|_p$ on $\mathbb{Q}$?

Question 3: If $|\cdot|$ is trivial on $\mathbb{Q}$, how to prove that $|\cdot|$ is trivial on $\mathbb{R}$?

Question 1 is easily answered as 'no' by considering $K = \Bbb{R}$, and let $|\cdot|$ be an extension of any $|\cdot|_p$ on $\Bbb{Q}$ into a valuation on $\Bbb{R}$ (existence of such an extension can be established using the Axiom of Choice). Since the restriction of $|\cdot|$ to $\Bbb{Q}$ is $|\cdot|_p$, it is non-trivial.

By the same reasoning (i.e. the example above), it must be impossible to provide the proof sought for in Question 2. Indeed, it is clear that the restriction to $\Bbb{Q}$ of an extension of $|\cdot|_p$ on $\Bbb{Q}$ is $|\cdot|_p$ itself.

No, the restriction of $|\cdot|$ to $\mathbb{R}$ need not be trivial. For instance, the valuation on $\mathbb{Q}_p$ extends to a valuation on its algebraic closure $\overline{\mathbb{Q}_p}$. We can choose a field isomorphism $\mathbb{C}\to\overline{\mathbb{Q}_p}$ to then get a valuation on $\mathbb{C}$ whose restriction to $\mathbb{Q}$ is the $p$-adic valuation.