Let $R$ be a Discrete Valuation Ring, and $K$ its fractions field. Now if $B\subseteq K$ is a subring with $R\subseteq B$ then we have $$B=R \text{ or } B=K.$$
Now this seems to be a very basic fact however I am just unable to prove it. Well after a hint from YACP: we can look at the at the uniformizer $p$ i.e. $v(p)=1$ - where $v$ is the valuation. We know for every element $\alpha \in K $ exists an $i\in \mathbb{Z}$ so that $ \alpha = u \cdot p^{i}$. Also we have that $K=S^{-1}R$ where $S=\{1,p,p^2,...\}$. I just can not say anything about $B\neq R$.:(
That $R$ is principal ideal domain with maximal ideal $m=(p)$ keeps coming up but I am unable to use it.