# Units in a discrete valuation ring.

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 $$x^{-1}\in \mathfrak m$$.