I am studying valuation theory at the moment and I came across this question:
Let $K$ be a field and let $a \in K$ such that $v(a) = 0$ for every valuation $v$ on $K$. What can we say about $a$?
In the positive characteristic case, for instance, we have the following:
Remark 1: Assume that $K$ has characteristic $p > 0$ and let $\mathbb F_p$ denote its prime field. Let $a \in K$. Then $v(a) = 0$ for every valuation $v$ on $K$ if and only if $a$ is algebraic over $\mathbb F_p$.
Proof of Remark 1: Assume first that $a$ is not algebraic over $\mathbb F_p$. Then $a$ is an irreducible element in the polynomial ring $\mathbb F_p[a]$ and hence induces a discrete rank-one valuation $w$ on its quotient field $\mathbb F_p(a)$ with $w(a) = 1$. This valuation $w$ extends to a valuation $v$ on $K$, because $\mathbb F_p(a) \subseteq K$. In particular, $v(a) = 1 \neq 0$. Now assume that $a$ is algebraic over $\mathbb F_p$. Then, clearly, $\mathbb F_p(a)$ is a finite field. It is well-known that finite fields do only admit the trivial valuation with constant value $0$. If now $v$ is a valuation on $K$, then $v$ restricted to $\mathbb F_p(a)$ is constant $0$ and in particular $v(a) = 0$, which ends the proof.
The first part of the above proof can be used to show also the following:
Remark 2: Let $K$ be any field and let $a \in K$ be transcendental over the prime field of $K$. Then there exists a valuation $v$ on $K$ such that $v(a) \neq 0$.
Example: If $K = \mathbb Q$ then by using $p$-adic valuations it is clear that there exists a valuation $v$ on $K$ with $v(a) \neq 0$ unless $a \in \{1,-1\}$.
Question: Let $K$ be of characteristic $0$ and let $a \in K$ be algebraic over $\mathbb Q$. What can we say about $a$?
Thank you in advance for your help!