Non-archimedean exponential valuation and integral closure I am trying to solve the following problem from Neukirch's book on ANT:  
Let  $L|K$ be a finite field extension, $v$ a nonarchimedean exponential valuation, and $w$ an extension to $L.$ If $\mathcal{O}_L$ is the integral closure of the valuation ring $\mathcal{O}_K$ of $v$ in $L$, then the localization $\mathcal{O}_{L,p}$ of $\mathcal{O}_L$ at the prime ideal $$p = \{ \alpha \in \mathcal{O}_L |  w(\alpha) >0 \}$$ is the valuation ring of $w.$  
So, my attempts so far is to take an $x = \alpha / \beta \in L$ where $\alpha,\beta \in \mathcal{O}_L.$ If we could show that we can choose $\beta$ such that $w(\beta) =0,$ I believe we would be done but I don't see why this should be true. Does anyone have an argument or some hint?
 A: It is easy to see that $\mathcal{O}_{L,p}$ is a subset of the valuation ring $\mathcal{O}_w.$ Call the maximal ideal of $\mathcal{O}_{L,p}$ $p$ as in your post. So, for the other direction consider an element $\alpha \in L$ such that $w(\alpha) \geq 0$ and write $\alpha= x/y.$ $\alpha$ then satisfies an equation with coefficients in $K,$ $$a_n(x/y)^n + \cdots +a_1(x/y)+a_0 = 0.$$ Choose $a_s$ with the smallest value $w(a_s)$ farthest to the left in the equation above. Set $b_m = a_m/a_s$ so that by dividing by $a_s$ we get an equation 
$$b_n(x/y)^n+\cdots b_1(x/y)+b_0=0.$$
We see that $b_n,\ldots,b_{s+1}$ all lie in $p$ while all $b_m$ lie in $\mathcal{O}_{L,p}.$ Dividing by $(y/x)^s$ we get an equation 
$$(b_n(x/y)^{n-s} + \cdots +b_{s+1}(x/y)+1)+y/x (b_{s-1}+ \cdots b_0 (y/x)^{s-1})=0.$$ Setting $$a =b_n(x/y)^{n-s} + \cdots +b_{s+1}(x/y)+1$$ and $$b = b_{s-1}+ \cdots b_0 (y/x)^{s-1}$$ we can rewrite the above equation as $a+by/x=0.$ Rearranging, we see that $x/y = -b/a$ so it is enough to show that $a$ and $b$ are in $\mathcal{O}$ and $a$ is not in $p.$ For this, we'll use that the integral closure of $\mathcal{O}_v$ in $L$ is the intersection of all valuation rings in $L$ lying above $\mathcal{O}_v$. If $x/y$ has non-negative valuation in some valuation ring $R$ lying above $\mathcal{O}_v$ then clearly $a$ has so as well, since it is a polynomial in $x/y$ and $b$ will also have non-negative valuation. Conversely, if $x/y$ has negative valuation, then $b$ has non-negative valuation since it is a polynomial in $y/x$ and one sees that $a$ then also has non-negative valuation.   
Lastly, one shows that $a$ is not in $p$ follows immedieately from $w$ being a non-archimedean valuation.
