Proposition 9.2 in Atiyah's commutative algebra book
I have read this related proof of proposition 9.2 (ii)=>(iii).
Here's the part of this proposition I need.
Proposition 9.2.
Let $A$ be a Noetherian local domain of dimension one, $\mathfrak{m}$ its
maximal ideal, $k = A/\mathfrak{m}$ its residue field. Then the following are equivalent:
ii) $A$ is integrally closed;
iii) $\mathfrak{m}$ is a principal ideal;
Here is the proof of this implication.
Let $a\in\mathfrak{m}$ and $a\neq 0$. By remark (A) there exists an integer $n$ such that $m^n\subset (a)$, $\mathfrak{m}^{n-1}\nsubseteq (a)$. Choose $b\in\mathfrak{m}^{n-1}$ 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 $x^{-1}\mathfrak{m}\nsubseteq\mathfrak{m}$(for if $x^{-1}\mathfrak{m}\subset\mathfrak{m}$, $\mathfrak{m}$ would be a faithful $A[x^{-1}]$-module, finitely generated as an $A-module$). But $x^{-1}\mathfrak{m}\subset A$ by construction of $x$, hence $x^{-1}\mathfrak{m}=A$ and therefore $\mathfrak{m}=Ax=(x)$.
Now it is clear why $x^{-1}\mathfrak{m}$ is an ideal and how one can define the set $x^{-1}\mathfrak{m}$.
The next question remains:
If $x^{-1}\mathfrak{m}\subset\mathfrak{m}$, then Why $\mathfrak{m}$ would be a faithful $A[x^{-1}]$-module, and finitely generated as an $A$-module? (I knew that the ideals in Noetherian ring are finitely generated.)
Best Answer
Answer to the first question. I use the notations and facts stated on this page of the book.
Since $b\in\mathfrak{m}^{n-1}$ and $x=a/b$ it follows that for any $v\in\mathfrak{m}$ we have $bv\in\mathfrak{m}^n\subset(a)$ and so $bv=ay$ for some $y\in A$ hence $$ x^{-1}v=\frac{bv}{a}=\frac{ay}{a}=y. $$
Supplement.
So we understand that $x^{-1}\mathfrak{m}\subset A$. Then since $\mathfrak{m}$ is an ideal, for any $z\in A$ we have $zx^{-1}\mathfrak{m}=x^{-1}(z\mathfrak{m})\subset x^{-1}\mathfrak{m}$ and $x^{-1}\mathfrak{m}+x^{-1}\mathfrak{m}=x^{-1}\mathfrak{m}$. It follows that $x^{-1}\mathfrak{m}$ is an ideal in $A$. Last, since $A$ is a local ring and $x^{-1}\mathfrak{m}\not\subset\mathfrak{m}$ it follows that $x^{-1}\mathfrak{m}=A$.
Supplement 2.
Since $K$ is the fraction field of ring $A$ and $\mathfrak{m}\subset A\subset K$, $x^{-1}\in K$, it follows that if $vt=0$ for $v\in\mathfrak{m}$ and $t\in A[x^{-1}]$, then either $v=0$ or $t=0$ and so $\mathfrak{m}$ is a faithful $A[x^{-1}]$-module.
Supplement 3.
I can't understand what's confusing you.
Here are the answers to your questions from your comment. Here is the logic of reasoning:
by choice $x^{-1}\notin A$ and therefore is not integral over $A$ (by convention ring $A$ is integrally closed);
we know that $\mathfrak{m}$ is an ideal in $A$ and if $x^{-1}\mathfrak{m}\subset\mathfrak{m}$, then $\mathfrak{m}$ is an $A[x^{-1}]$-module. Since $\mathfrak{m}$ is finitely generated as an $A$-module ($A$ is a Noetherian ring) and $\mathfrak{m}$ is a faithful $A[x^{-1}]$-module ($A$ is an integral domain), it follows from Proposition 5.1 that $x^{-1}$ is integral over $A$. Contradiction. Hence we conclude that $x^{-1}\mathfrak{m}\not\subset\mathfrak{m}$.
Why if $x^{-1}\mathfrak{m}\subset\mathfrak{m}$, then $\mathfrak{m}$ is an $A[x^{-1}]$-module?
By the definition of module, it is sufficient to check only $tv\in\mathfrak{m}$ for any $v\in\mathfrak{m}$ and $t\in A[x^{-1}]$. It follows from the fact that $x^{-1}\mathfrak{m}\subset\mathfrak{m}$ and ring $A[x^{-1}]$ is generated by ring $A$ and $x^{-1}$, i.e. the elements in $A[x^{−1}]$ are polynomials of $x^{−1}$ with coefficients from $A$.
$x^{-1}\mathfrak{m}\subset A$ by construction of $x$;
since $A$ is local ring and $\mathfrak{m}$ its only maximal ideal and $x^{-1}\mathfrak{m}$ is also an ideal not lying in $\mathfrak{m}$, it follows that $x^{-1}\mathfrak{m}=A$ and $\mathfrak{m}=xA=(x)$.
Supplement 4.
Answers to the questions in the comment below
Correct.
I don't understand this question. On the contrary $\mathfrak{m}^n\subset(a)$. Perhaps you are asking why $\mathfrak{m}^{n-1}\not\subset(a)$? If $\mathfrak{m}\subset(a)$, then $\mathfrak{m}=(a)$ ($\mathfrak{m}$ is a maximal ideal in $A$). So $\mathfrak{m}$ is a principal ideal with generating $a$. This is what we are proving. If $\mathfrak{m}\not\subset(a)$, then there exists such $n$ that 2 conditions $\mathfrak{m}^n\subset(a)$ and $\mathfrak{m}^{n-1}\not\subset(a)$ simultaneously hold.
Absolutely not. See (iv) of Proposition 5.1. The role of $M$ is played by the ideal $\mathfrak{m}$.
The part of Proposition 5.1 that we need can be formulated like this: