The ring of integers of a number field is contained in any valuation ring. Let $K$ be a number field and let $v$ be a surjective discrete valuation on $K$.  Let $\mathcal O_v$ the valuation ring of $v$, namely 
$$\mathcal O_v=\{x\in K\,: v(x)\ge 0\}$$
Let's denote with $O_K$ the ring of integers of $K$. Why do we have the inclusion $O_K\subset\mathcal O_v$.
I other word, why for any $y\in O_K$, then $v(y)\ge 0$?
Many thanks in advance 
 A: Hint: Show that $\mathcal{O}_v$ is integrally closed and contains $\mathbb{Z}$ (cf., Proposition 5.18 of Atiyah and MacDonald), and recall that $O_K$ is the integral closure of $\mathbb{Z}$ in $K$.
A: You can also show it directly: if $\alpha\in \mathcal O_K$, then $\alpha^n+\sum_{i=1}^{n-1}b_i\alpha^i=b_0$ where $b_0,\ldots,b_{n-1}\in\mathbb Z$ and $b_0\neq 0$. If $v(\alpha)<0$, then for every $i=1,\ldots,n-1$ one has that $v(b_i\alpha^i)=v(b_i)+iv(\alpha)>nv(\alpha)=v(\alpha^n)$ because $v(x)\geq 0$ for every $x \in \mathbb Z$. Therefore $v(\alpha^n)\neq v\left(\sum_{i=1}^{n-1}b_i\alpha^i\right)$, which proves that $v\left(\alpha^n+\sum_{i=1}^{n-1}b_i\alpha^i\right)=v(\alpha^n)<0$, contradiction since $v(b_0)\geq 0$.
A: Another simple proof : if $\alpha \in K$ is integral over $\mathbf Z$, then a fortiori $\alpha \in K_v$ is integral over $\mathbf Z_p$. But the integral closure of $\mathbf Z_p$ in $K_v$ coincides with the valuation ring (= $O_v$ in your notation) because the conjugates of an element of $K_v$ have the same valuation, see https://math.stackexchange.com/a/2024567/300700 (NB:the notations are not the same there)
