Let $(A,\mathfrak{m})$ be a Noetherian local domain of dimension one that is also integrally closed (i.e. a local Dedekind domain), then $\mathfrak{m}$ is a principal ideal.
I am trying to follow the proof of Atiyah's Introduction to Commutative Algebra in proposition 9.2. $ii)\Rightarrow iii)$.
The proof is as follows:
Let $a\in\mathfrak{m}$. There exists $n\in \mathbb{N}$ such that $\mathfrak{m}^{n}\subseteq (a)$ but $\mathfrak{m}^{n-1}\nsubseteq(a) $. Choose $b\in\mathfrak{m}^{n-1}-(a)$. Then, if $x=a/b$, we have that $x^{-1}\notin A$ and therefore $x^{-1}$ is not integral over $A$. Now Atiyah says that $x^{-1}\mathfrak{m}\nsubseteq\mathfrak{m}$. Why is this true?
According to the book, if $x^{-1}\mathfrak{m}\subseteq\mathfrak{m}$, then $\mathfrak{m}$ would be a faithful $A[x^{-1}]$-module, finitely generated as an $A$-module.
I don't see why $\mathfrak{m}$ is finitely generated as an $A$-module, and I don't understand why this would be a contradiction.
$\mathfrak m$ is finitely generated as an ideal or as an $A$-module because $A$ is noetherian.
From Proposition 5.1(iv) we get that $x^{-1}$ is integral over $A$, a contradiction.