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.

  • $\begingroup$ Instead of $A$ did you mean $R$? if not, what is $A$?! $\endgroup$ – rschwieb May 14 '13 at 19:57
  • $\begingroup$ Did you mean somewhere that $\,A= R\,$ ...? $\endgroup$ – DonAntonio May 14 '13 at 19:58
  • $\begingroup$ Take an element $z$ of $B \setminus R$; think about what it means to be a fraction field. Use the form $u p^i$ just mentioned. Then show that every element of $K$ is of the form $r z^n$, for $r\in R$ and $n$ a nonnegative integer. $\endgroup$ – neilme May 14 '13 at 22:33

As $R$ contains exactly the elments $a\in K$ with $|a|\le 1$, the ring $B$ must contain some element $|b|$ with $|b|>1$. If $x\in K$ is arbitrary, therefore $|x|<|b|^n$ for some $n$ and the element $\frac{x}{b^n}$ is in $R$, hence $x=\frac x{b^n}\cdot b^n$ is in $B$.


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