Problem 2.26 in Fulton's Algebraic Curves: redundant hypothesis? The problem reads: "Let $R$ and $S$ be DVRs with maximal ideals $M = (q)$ and $N = (p)$ respectively, $K$ the quotient field of $R$. Suppose $R \subset S \subset K$, and suppose that $M \subset N$. Then show that $R = S$."
The following proof is not making use of the fact that $S$ is a DVR.
"Proof" Suppose there is some $s \in S \setminus R$. Then $s = uq^n$, for $n < 0$, so $s^{-1} = uq^{-n} \in M \subset N$. $s \in S$, so $sN \subset N$, but $s^{-1} \in N$ now implies that $N = (1)$, which contradicts properness of $N$.
Is there an error in the proof or a more general statement holds?
 A: The essence of your question is the following: if $R$ is a DVR with field of fractions $K$, then can there exist a ring $S$  that is not a DVR such that $R \subset S \subset K$ and the maximal ideal of $R$ is inside a maximal ideal of $S$?
I believe the answer to this question is no. First notice that $S$ must be an integral domain by the inclusion $S \subset K$. Second, $K$ must also be 
the field of fractions of $S$. Third, if $x \in K - S$, then $x \in K - R$ and so $x^{-1} \in R$, which implies $x^{-1} \in S$. This proves that $S$ must be a valuation ring and thus a local domain.
So it seems to me that a better stated exercise would be:
"Let $R$ be a DVR with field of fractions $K$ and let $S$ be a ring such that
$R \subset S \subset K$. Then show that i) $S$ is a valuation ring ii) if 
the maximal ideal of $R$ is inside the maximal ideal of $S$, then $R=S$."
To conclude, your proof is correct and it is a proof of a more general statement.
A: I think the condition that maximal ideal of $R$ should be contained in the maximal ideal of $S$ is not needed as follows : since $R$ is a DVR , therefore  $ R = \{ u{t}^n $ and $n\geqslant 0$ : $u$ is a unit in $R$ and $t$ is a uniformizing parameter of $R$ } and $ K = \{ u{t}^n : n \in \ Z $ and  $u$ is a unit in $R$ }. So suppose if $ R \subsetneq S $ then $ \exists \ x \in S-R$ which means $ x = u{t}^a$ such that $ a<0$. Now since $t \in R\subset S$ which means $ t^{-1} \in S$ after multiplying $x$ by a suitable power of $t\in S$. Hence, $S=K$ as K is generated as a ring from $t , \ t^{-1}$ and all the units in $R$. Thus $S$ is a field which are not DVRs which is a contradiction. So , whenever $R\subset S$ are DVRs with same quotient field then $R=S$.
