Consider a normal Noetherian local ring $(A,\mathfrak m)$ of dimension $1$. I am working through a proof that such a ring is a principal ideal domain.
Consider $x\in \mathfrak m \backslash \mathfrak m^2$. Such an element exists, otherwise the ring would have dimension $0$. By the theorem that $\dim(A/(f))\ge \dim A -1$ for any $f$, we see that $\dim A/xA$ is $0$. Why does this imply that $\mathfrak m = \sqrt{xA}$? And why does it also imply that $\mathfrak m^n\subset xA$ for some $n$? Certainly for every $m\in\mathfrak m$, some power must lie in $xA$, but why can't the supremum of these powers be infinite?
If $A/xA$ has dimension zero, then it is an Artin local ring (because $A$ is Noetherian), so its maximal ideal, which is $\mathfrak{m}/xA$, is nilpotent (this is true for any Artin local ring), meaning $\mathfrak{m}^n\subseteq xA$ for some $n\geq 1$. This implies that $\mathfrak{m}\subseteq\sqrt{xA}$, and since $xA$ is a proper ideal, $\mathfrak{m}=\sqrt{xA}$.