How to show that any nontrivial valuation on the field of rational numbers is equivalent to some p-adic valuation?

Question

I have started with a nontrivial valuation v on Q. Suppose V is the corresponding valuation ring. I want to show that V is equal to the valuation ring of some p-adic valuation for some prime p. Then by definition of equivalence this will hold. For this I use the fact that as v is nontrivial valuation so there must exist atleast one prime p such that 1/p is not in V. From this I can show that V⊆Vp where Vp is the valuation ring of the p-adic valuation on Q which is precisely the set of rationals with denominator not divisible by p. But am stuck with the other way inclusion. Can anybody suggest to proceed furthur?

Answer 1

I don't know how you obtained $O_v\subset \Bbb{Z}_{(p)}$.

$O_v\subsetneq \Bbb{Q}$.

$v(1)=0$ gives that $v(n)\ge 0$ thus $\Bbb{Z}\subset O_v$.

There is some $p$ such that $v(p)>0$ because otherwise $O_v=\Bbb{Q}$.

$m_v\cap \Bbb{Z}$ is a prime ideal containing $p$ thus it is $(p)$ and hence $O_v$ contains $\Bbb{Z}_{(p)}$.

For any $a\in \Bbb{Q}-\Bbb{Z}_{(p)}$, $\Bbb{Z}_{(p)}[a]=\Bbb{Q}$, thus $O_v=\Bbb{Z}_{(p)}$.