While revising for my exam in Commutative algebra, I came across the following statement that:
Let $A$ be a Noetherian local domain of dimension one. Let $x,y \in A$ with $x \neq 0$ and $y \in \mathfrak{m}$. Then $ax = y^n$ for some $a \in A$ and $n \geq 1$.
But I don't really understand why? Does this have anything to do with that in a Noetherian local domain every non-zero ideal can be written as a power of the maximal ideal? If yes, why do we even need an $a \in A$?
Thanks in advance!
If $A$ is a DVR, then every nonzero ideal in a DVR is a power of the maximal ideal $\mathfrak{m}$, and we have $x=ut^n$ for $t\in \mathfrak{m}$ being a uniformizer. However, if $A$ is not a DVR, then this is not true in general. Then $\mathfrak{m}$ is not principal.