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Let $S$ be a discrete valuation ring and $R\subset S$ be a proper subring (also a DVR). Assuming that $M$ and $N$ are the respective maximal ideals of $R$ and $S$ and that $N\cap R = M$, then the quotient field of $S$ is a proper subfield of the quotient field of $R$.

My Idea: Take $x\in S - R$ which must be non-zero. Then $(x)$ is an ideal of $S$. Since $(x)\neq 0$ and $S$ is a discrete valuation ring, then we must have that $(x)$ is an ideal of $S$.

I know this doesn't go very far, but there are a few hypothesis that are bizarre to me.

The requirement that $N\cap R = M$ sounds a bit like the requirements of the "going up" theorem [of which there seems to be quite a few versions :(] but since $R$ is integrally closed, $S$ cannot be integral over $R$. So this is paragraph I have all but completely dismissed.

To complete the story, I am stuck.

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    Yep! That's correct.2012-03-17

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I think you are implicitly assuming that $R$ is also a DVR, so I will also use this.

Assume the fraction fields of $R$ and $S$ are equal: $R\subseteq S\subseteq \text{Frac}(R)$. Let $0\neq a\in S$, we want to show that $a\in R$: If $v_R(a)\geq 0$ we're done, so assume $v_R(a)<0$, which implies $v_R(a^{-1})>0$ and so $a^{-1}\in M\subseteq N$, but that is impossible as $a\in S\Rightarrow a^{-1}\notin N$.

We have shown that all $a\in S $ satisfy $v_R(a)\geq 0$ and so $R=S$.

EDIT: In fact it is enough to assume that $R$ and $S$ are valuation rings.

Assume the fraction fields of $R$ and $S$ are equal. Take $a\in S\backslash R$. Since $R$ is a valuation ring $a^{-1}\in R$, but $a^{-1}$ is not a unit in $R$ and hence is in the maximal ideal $M$ of $R$. $a^{-1}\in M\subseteq N\subseteq S$. This is a contradiction, as $a^{-1}\in N$ implies that $a^{-1}$ is not a unit in $S$ and hence $a\notin S$.

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    No need to be sorry. You've laid this out in a very basic way. I should be able to follow it immediately. The only thing slowing me down is that we were given a different definition of a DVR, so in order to use your argument, I first must prove the equivalent of definitions (which a peer of mine says is possible with a couple of theorems we proved in class, thankfully). To clarify: me not yet checking the answer in no way means that I don't accept/believe/trust it. I just have to catch up. :)2012-03-18