How does it follows that A normal local ring of altitude 1 is a valuation ring (12.3) Let $P$ be a prime ideal of Noetherian ring $R$. Assume $R_P$ is not a valuation ring. Then for any non-zero divisor $a\in P$ and $b\in (aR:P)_R$, $b/a$ is integral over $R$ and the conductor $(R:R[b/a])_R$ contains $P$.
(12.4) Theorem: A normal local ring $(R,m)$ of altitude 1 is a valuation ring. 
$\textbf{Q:}$ How does 12.4 follows immediately from 12.3? In particular, how does 12.3 implies that maximal ideal $m$ is principally generated? The following is my consideration. Take $a\in m-m^2$. Clearly $0\neq b\in (aR:m)_R$. I want to see $(aR:m)=R$ which will implies that $m$ is principally generated. Assume $(aR:m)\subset m$. Thus we have a $b\in m\cap(aR:m)_R$ s.t. $b/a\in R$ by $R$ normal via 12.3. How should I proceed?
Ref. Nagata, Local Rings, Chpt 1, Sec 12.
 A: Let $(R, \mathfrak{m})$ be a $1$-dimensional normal local ring.  (Local here including Noetherian). Note that this implies $R$ is domain, so we can forget about non zero divisor conditions in the statement of $12.3$.    
To my mind we need to milk the '$1$-dimensional local' hypotheses a little bit more to get from $12.3$ to $12.4$.  In particular, if you combine the (minimal) primary decompositions you get from Noetherianity with the fact that there's only one nonzero prime ideal, then you get that every ideal of a $1$-dimensional local ring $(R, \mathfrak{m})$ is $\mathfrak{m}$-primary.  Thus if $a$ is any element of $\mathfrak{m}$ then $\sqrt{aR} = \mathfrak{m}$ and (using that $\mathfrak{m}$ is f.g.) you have $\mathfrak{m}^n \subseteq aR$ for some $n$.  Nagata covers primary decompositions in sections $7$ to $8$, if that's your reference.
From the existence of $a \in \mathfrak{m}$ such that $\mathfrak{m}^n \subseteq aR$, the implication $(12.3 \implies 12.4)$ is really immediate.  Indeed, if $R$ wasn't a valuation ring, then you could iteratively apply $12.3$ to get $\mathfrak{m}^{n-i} 
\subseteq aR$ for $0 \leq i < n$.....
