Let $R=k[x,y]_{(x,y)}$ and $f\in R$. My question is:
What is the necessary and sufficient condition for $R/(f)$ to be a Dedekind domain?
My guess is that $R/(f)$ is a Dedekind domain if and only if $(x,y)^2\subsetneq J\subsetneq (x,y)$, where $J=(f)\cap k[x,y]$.
I proved that it is a sufficient condition in the following way:
Clearly $(f)$ is a height $1$ prime ideal. Hence $J$ is also a height $1$ prime ideal. Since $k[x,y]$ is a Noetherian UFD $J=(h)$, where $h$ is a polynomial in two variables with zero as constant term.
If $h\notin (x,y)^2$ and $h\in (x,y)$, we can write: $$h=xa(x,y)+yb(x,y),$$ where either $a(x,y)\notin (x,y)$ or $b(x,y)\notin (x,y)$. Without loss of generality let $a(x,y)\notin (x,y)$. In that case in $(k[x,y]/(h))_{\overline{(x,y)}}$ we have $\overline{x}=-\overline{y}\frac{\overline{b(x,y)}}{\overline{a(x,y)}}$. Therefore $\overline{(x,y)}=\overline{y}$ in $R/(f)$.
Clearly $R/(f)$ is a Noetherian local ring with $\dim>0$. Also above we proved that the maximal ideal in $R/(f)$ is principal. Hence $R/(f)$ is a DVR and consequently it is a Dedekind domain.
For the necessary condition:
Let $R/(f)$ be a Dedekind domain. A Dedekind domain is either a field or a Noetherian one dimensional normal domain. Now if $R/(f)$ is a field then $(f)=(x,y)k[x,y]_{(x,y)}$ as $R$ is a local ring. Then Krull's Principal ideal theorem height of $(x,y)k[x,y]_{(x,y)}$ is less or equal to $1$ which is a contradiction as we know height of this prime ideal is $2$.
Clearly, $h\in (x,y)$. How do I show that $h\notin (x,y)^2$?