I'm struggling with the following question: suppose that $K$ is a Dedekind Domain with ring of integers $\mathcal{O}_K$, and that we have an element $x\in\mathfrak{p}$ such that $x\not\in\mathfrak{p}^2$, where $\mathfrak{p}\subset\mathcal{O}_K$ is a prime ideal. Let $y\not\in\mathfrak{p}$. Is $xy$ in $\mathfrak{p}^2$?
I think I am supposed to use localization; would the argument run as follows?
Let $\mathcal{O}_{K_\mathfrak{p}}$ denote the localization of $\mathcal{O}_K$ at $\mathfrak{p}$. Then for all $x\in\mathcal{O}_K$, the map $f:x\mapsto\frac{x}{1}$ is an injective homomorphism. Suppose $xy$ is in $\mathfrak{p}^2$. Then $\frac{xy}{1}\in\mathfrak{p}^{-1}\mathfrak{p}^2$ (the ideal in $\mathcal{O}_{K_\mathfrak{p}}$ generated by $f(\mathfrak{p}^2)$). But then we also have $\frac{xy}{y}\in\mathfrak{p}^{-1}\mathfrak{p}^2$, so $\frac{x}{1}\in\mathfrak{p}^{-1}\mathfrak{p}^2$. But then $x$ is in the preimage of $f(\mathfrak{p}^2)$, so we would have $x\in\mathfrak{p}^2$ - which is a contradiction.