It was a exercise that one of our professors gave it to us and I don't know the solution of it:
Suppose that $v$ is a non-archimedean valuation on the field $F$ and $o(v)$ is the valuation ring of $v$. (It means $o(v)=\{x \in F : v(x)\leq1\}$.) If $x\in F$ is algebraic over $o(v)$, then prove that $x$ is in $o(v)$.
A non-archimedean valuation has the following property: $v(a+b)=\max(v(a),v(b))$ if $v(a)\neq v(b)$. If $x\in F$ is integral over $O_v$ then there exist $a_i\in O_v$ such that $\sum_{i=0}^{n-1}a_ix^i+x^n=0$. Suppose $v(x)>1$. Then $v(a_ix^i)<v(x^n)$ for all $i=0,1,\dots,n-1$ since $v(a_i)\le 1<v(x)$, and therefore $v(\sum_{i=0}^{n-1}a_ix^i+x^n)=v(x^n)>0$, a contradiction.