Let $K$ be a field.
By a discrete valuation on $K$, we mean a valuation whose value group has a minimal element $>0$. For example, the value group is $\mathbb Z \times \mathbb Z$ with lexicographic order.
By a discrete rank 1 valuation on $K$, we mean a valuation whose value group is isomorphic to $\mathbb Z$.
I am wondering, if there exists a discrete rank 1 valuation $v$ on $K$, is there always a discrete valuation $w$ on $K$ such that $v$ and $w$ are independent (in the sense that they do not share a prime ideal or, equivalently, that they generate $K$ as a ring).
Of course, global fields have that property, since they have infinitely many disrete rank 1 valuations. Also, if $K$ has this property, then every purely transcendental extension has it.
A possible counterexample could be $\mathbb Q_p$, but I am not sure.
Thank you for your help!
Let $v_p$ be the canonical valuation.
If $w(a)=0$ whenever $a\in 1+p \Bbb{Z}_p$ then $w(a)=0$ whenever $a\in \Bbb{Z}_p^\times$ so that $\forall a\in \Bbb{Q}_p^\times,w(a) = w(p)v_p(a)$.
$w(1)=0$ gives that $w(p)\ge 0$ and hence $w$ is either trivial or equivalent to $v_p$.
Whence, that $w$ is non-trivial and non-equivalent to $v_p$ implies that $w(a) \ne 0$ for some $a\in 1+p\Bbb{Z}_p$.
By Hensel lemma, for all $p\nmid n$, $a^{1/n}\in 1+p \Bbb{Z}_p$ so that $w(1+p\Bbb{Z}_p)$ contains $\Bbb{Z}_{(p)} w(a)$ which contradicts that $w$ is discrete.