Let be $K$ a field and $v:K^{*}\longrightarrow \mathbb{Z}$ a discrete valuation. We can suppose $v$ is sobrejective.
The set $R=\{x\in K^{*}:v(x)\geq 0\}\cup \{0\}$ is called the valuation ring of $v$.
How can i show $q.f(R)=K$?
I wrote a possible answer but i am not sure:
Firts, $v(1)=0$ because $v(1)=v(1*1)=v(1)+v(1)\Rightarrow 0=v(1)$.
$\subset]$ $R\subset K \Rightarrow q.f(R)\subset q.f(K)=K$
$\supset]$ We take an arbitrary $x\in K^{*}$.
If $v(x)\geq 0$ then $x\in R$, so $x=\frac{r_1}{r_2}$ with $r_1:=x\in R$ and $r_2:=1\in R$ (because $v(1)=0$).
If $v(x)<0$, then $\exists n\in\mathbb{N}:v(x)=-n$. We know $v$ is sobrejective, so i can take $r_2\in K^{*}:v(r_2)=n+1$, so $r_2\in R$. Now i define $r_1=xr_2$, then $v(r_1)=v(xr_2)=v(x)+v(r_2)=-n+n+1=1>0$, then $r_1\in R$ and $x=r_1/r_2$.
Is this proof correct?
I think your proof works but it is a little harder than necessary for the second inclusion.
Given $x\in K^*$, either $x\in R$ (when the valuation is nonnegative) or $1/x \in R$ because $v(1/x) = -v(x) \geq 1$. In the first case, obviously $x\in R \subseteq Frac(R)$ and in the second, $x= 1/(1/x)$ is in $Frac(R)$.