I'm trying to work out a criterion that $R[\sqrt{D}]$ be a Dedekind ring.
Suppose the $R$ is integrally closed in its quotient field $F$, $2$ is a unit in $R$, and let $F(\sqrt{D})$ be a quadratic extension for some $D\in R$ nonzero.
I have been able to show that $x+y\sqrt{D}\in F(\sqrt{D})$ is integral over $R$ iff $x,Dy^2\in R$.
Now suppose $R$ is Dedekind, $2$ is a unit in $R$, and $(D)$ is not divisible by the square of a prime ideal in $R$. Apparently it follows in this case that $R[\sqrt{D}]$ is Dedekind.
Why? My hunch is that $R[\sqrt{D}]$ is the integral closure of $R$ in $F(\sqrt{D})$, in which case $R[\sqrt{D}]$ would be a Dedeking ring by a well known theorem.
By the above criterion, its clear that $R[\sqrt{D}]$ is integral over $R$. On the other hand, suppose $x+y\sqrt{D}\in F(\sqrt{D})$ is integral over $R$. Again by the above criterion, $x\in R$, and $Dy^2\in R$. It is enough to show $y\in R$.
The fact that $Dy^2\in R$ and $(D)$ is not divisible by the square of a prime ideal seem like they should be related to show $y\in R$, but I'm not sure of the link here. Since $D\mid Dy^2$ and $y^2\mid Dy^2$ in $R$, and $R$ is Dedekind, there exist ideals $I$ and $J$ such that $(Dy^2)=(d)I=(y^2)J$. I could further factor these into prime ideals, but that's it.
If $y$ is not in $R$, then since $R$ is Dedekind, there is a nonzero prime ideal $\mathfrak{p}$ of $R$ such that $v_{\mathfrak{p}}(y)$ -- the exponent of $\mathfrak{p}$ in the prime-power factorization of the fractional ideal $(y)$ -- is negative. You want to play that off against the condition that $D$ is not divisible by the square of any prime ideal.
If you don't see how to do this, try it in $\mathbb{Z}$ (in which $2$ is not a unit, but never mind that) first: if $y \in \mathbb{Q}$, $D$ is a squarefree integer, and $Dy^2 \in \mathbb{Z}$, then show that $y \in \mathbb{Z}$. The case of a general Dedekind domain is very similar.