I'm doing problem 2.26 in the book "algebraic curves" by Fulton:
http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf
One is given two DVR's R and S both of which have the same field of fractions $K$. It is assumed that the maximal ideal of $R$ is contained in that of $S$, and the problem is to show that $R$ and $S$ are equal.
I managed to show that if an element of $S$ is in the maximal ideal of $S$ but not in the maximal ideal of $R$, then its inverse is also in the maximal ideal of $S$, which implies that $S=K$, a contradiction (as a DVR may not be a field).
My question is then, having showed that $R, S$ must have the same maximal ideal, how does it follow that $R=S$, i.e., what about the units?
Let $x\in S$. Then, either $x\in R$, and there's nothing to prove, or $x\notin R$. This implies by the 1st question of problem 2.26 that $x^{-1}\in\mathfrak m$ (the maximal ideal common to $R$ and $S$). As $\mathfrak m$ is also the maximal ideal of $S$, this means that both $x$ and its inverse are is $S$ – in other words, $x$ is a unit in $S$. This is impossible since $x^{-1}\in \mathfrak m$.