# Discrete valuations of the rational numbers

I'm trying to find every discrete valuation on the field of rational numbers.

If $a\in \mathbb Q$, we can write $a=p^j\frac{x}{y}$, where $p$ is a prime number and $p\nmid x$ and $p\nmid y$. We can define the p-adic valuation $v_p(a)=j$.

I found easy to prove this is indeed a discrete valuation. I would like to know if there is another discrete valuation over the rational numbers and if the answer is yes, how many? why?

I need help

Given a discrete valuation $\nu:\mathbb{Q}^\times\to\mathbb{Z}$, we have $\nu(1)=0$, so $\nu(n)=\nu(1+1+\cdots +1) \geq 0$ for all $n\in\mathbb{N}$ (hence all $n\in\mathbb{Z}$).
Let $P=\{n\in\mathbb{\mathbb{Z}} \mid \nu(n)>0\}$. Then it is not hard to show that $P$ is a prime ideal of $\mathbb{Z}$, so $P=(p)$ for some prime number $p$. It follows that, if $(p,k)=1$, we have $\nu(p^j k) = \nu (p) + \nu(p) + \cdots + \nu(p) + \nu(k) = j\cdot \nu(p)$. We can easily extend this to all rationals, so that $\nu$ equals some multiple of the standard $p$-adic valuation.
• @user85493 That isn't true; any multiple of a $p$-adic valuation also works. If you want more details, please specify what you're looking for, as I provided 90% of the proof that these are the only possibilities. Aug 7, 2014 at 3:19
• I didn't understand why $\nu(p^jk)=j\cdot \nu(p)$ implies $\nu$ is equal of some multiple of $\nu_p$, where $\nu_p$ is the p-adic valuation. Aug 7, 2014 at 14:03
• @user85493 The $p$-adic valuation is $\nu_p(p^j k) = j$. So we have $\nu = \nu_p \cdot c$, where $c=\nu(p)$. Aug 7, 2014 at 15:42
• @user85493 Oh, it occurs to me that I forgot the case $P=(0)$. In that case, $\nu \equiv 0$ is the trivial valuation (which is sometimes, rarely, included in the definition of a discrete valuation). Aug 7, 2014 at 22:07
For further background reading: valuations correspond to nonarchimedean places, and there is a theorem of Ostrowksi which classifies all places $\Bbb Q$, and this generalizes to all global fields.