5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Instead of $A$ did you mean $R$? if not, what is $A$?! $\endgroup$
    – rschwieb
    May 14, 2013 at 19:57
  • $\begingroup$ Did you mean somewhere that $\,A= R\,$ ...? $\endgroup$
    – DonAntonio
    May 14, 2013 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, 2013 at 22:33

1 Answer 1

Reset to default
3
$\begingroup$

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 $B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.