Proposition 9.2. Let $A$ be a Noetherian local domain of dimension one, $m$ its maximal ideal,$k=A/m$ its residue field. Then the following are equivalent :
ii) $A$ is integrally closed;
iii) $m$ is a principal ideal;
Proof. Let $a\in m$ and $a\neq 0$. By remark (A) there exists an integer $n$ such that $m^n\subset (a)$, $m^{n-1}\nsubseteq (a)$. Choose $b\in m^{n-1}$ and $b\notin (a)$, and let $x=a/b\in K,$ the fraction field of $A$. We have $x^{-1}\notin A$(since $b\notin (a)$),hence $x^{-1}$ is not integral over $A$, and therefore by (5.1) we have $x^{-1}m\nsubseteq m$(for if $x^{-1}m\subset m$, m would be a faithful A[x^{-1}]-module, finitely generated as an $A-module$). But $x^{-1}m\subset A$ by construction of $x$ ,hence $x^{-1}m=A$ and therefore $m=Ax=(x)$.
The proof puzzles me is that we got $x^{-1}m \subset A$,then $x^{-1}m = A$. Why is that $x^{-1}m = A$ ? May I need to show that $m \subsetneqq x^{-1}m$ (But I think this is diffcult for me.Could you give me some hints), then the conclusion is followed by that $m$ is a maximal ideal?