 Solution

Choose $0\ne a\in \mathfrak{m}$. Since the one-dimensional local domain has no nonzero prime ideal other than $\mathfrak{m}$, one has $\sqrt{(a)}=\mathfrak{m}$. Finite generation of $\mathfrak{m}$ therefore gives some $n\ge 1$ with ${\mathfrak{m}}^n\subseteq (a)$. Choose $n$ minimal andIn the fraction field of $A$, put $y=x/a$. Then $y\notin A$ but

If $y\mathfrak{m}\subseteq \mathfrak{m}$, multiplication by $y$ would preserve the nonzero finitely generated faithful $A$-module $\mathfrak{m}$. The determinant trick would make $y$ integral over $A$, contradicting that $A$ is integrally closed and $y\notin A$. Hence some $t\in \mathfrak{m}$ satisfies $yt\notin \mathfrak{m}$. Since $yt\in A$ and $A$ is local, $yt$ is a unit.

For any $z\in \mathfrak{m}$, one has $yz\in A$, and thereforeThus $\mathfrak{m}\subseteq (t)$, while the reverse inclusion follows from $t\in \mathfrak{m}$. ConsequentlyThis proves the principal maximal ideal in a one-dimensional normal local domain result.

Solved by gpt-5.6-sol high.