An integral domain $A$ is a discrete valuation ring if there is a discrete valuation $v$ of its field of fractions $K$ such that $A$ is the valuation of $v$. By (5.18), $A$ is a local ring, and its maximal ideal of $m$ is the set of all $x \in K$ such that $v(x)＞0$.
We are going to prove that the set $S=\{x |x\in K,v(x)＞0\}∪\{0\}$ is a maximal ideal of $A$.
Claim 1.$S$ is an ideal of $A$.
Proof of claim 1: $A$ is valuation of ring of $v$. Then $v(x)≥0$ for all $x \in A$ .$s\in S, t\in A,v(s-t)≥min(v(s),v(-t))=min(v(s),v(t)) ＞0$ $v(st)=v(s)+v(t)＞0$. Thus $S$ is an ideal of $A$.
Claim 2. $S$ is a maximal ideal of $A$.
Proof of claim 2:$1\notin S $ $ (v(1)=0)$ , any ideal $B\subset A$ and $S \subsetneqq B$, then $B=A$. We only need to show that for some $r\in A, v(r)=0$, then $r$ is unit in $A$. But how to find the inverse of $r$?
