By a result that I know by the name of the "Kummer-Dedekind theorem" (see Chapter 3 of these notes), there is a unique prime ideal $\mathfrak{p}=(2,1+\sqrt{3})$ of $R = \mathbb{Z}[\sqrt{-3}]$ lying over $2$. By the same result, the rational prime $2$ is inert in $\widetilde{R} = \mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, i.e. $\mathfrak{q}=(2)$ is a prime ideal in $\widetilde{R}$. So for the corresponding map of prime spectra $\operatorname{Spec}(\widetilde{R})\rightarrow \operatorname{Spec}(R)$, the preimage of the "singular" point $\mathfrak{p}$ of $\operatorname{Spec}(R)$ consists solely of the "regular" point $\mathfrak{q}$ of $\operatorname{Spec}(\widetilde{R})$. I would therefore say that this situation is more like that of the cuspidal point $(0,0)$ on the singular curve $y^2=x^3$, which also has one point lying over it on the normalization.

By a result that I know by the name of the "Kummer-Dedekind theorem" (see Chapter 3 of these notes), there is a unique prime ideal $\mathfrak{p}=(2,1+\sqrt{3})$ of $R = \mathbb{Z}[\sqrt{-3}]$ lying over $2$. By the same result, the rational prime $2$ is inert in $\widetilde{R} = \mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, i.e. $\mathfrak{q}=(2)$ is a prime ideal in $\widetilde{R}$. So for the corresponding map of prime spectra $\operatorname{Spec}(\widetilde{R})\rightarrow \operatorname{Spec}(R)$, the preimage of the "singular" point $\mathfrak{p}$ of $\operatorname{Spec}(R)$ consists solely of the "regular" point $\mathfrak{q}$ of $\operatorname{Spec}(\widetilde{R})$. I would therefore say that, at least in this sense, the situation is more like that of the cuspidal point $(0,0)$ on the singular curve $y^2=x^3$, which also has one point lying over it on the normalization.

By a result that I know by the name of the "Kummer-Dedekind theorem" (see Chapter 3 of these notes), there is a unique prime ideal $\mathfrak{p}=(2,1+\sqrt{3})$ of $R = \mathbb{Z}[\sqrt{-3}]$ lying over $2$. By the same result, the rational prime $2$ is inert in $\widetilde{R} = \mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, i.e. $\mathfrak{q}=(2)$ is a prime ideal in $\widetilde{R}$. So the corresponding map of prime spectra $\operatorname{Spec}(\widetilde{R})\rightarrow \operatorname{Spec}(R)$ maps the "regular" point $\mathfrak{q}$ of $\operatorname{Spec}(\widetilde{R})$ to the "singular" point $\mathfrak{p}$ of $\operatorname{Spec}(R)$. I would therefore say that this situation is more like that of the cuspidal point $(0,0)$ on the singular curve $y^2=x^3$, which also has one point lying over it on the normalization.

By a result that I know by the name of the "Kummer-Dedekind theorem" (see Chapter 3 of these notes), there is a unique prime ideal $\mathfrak{p}=(2,1+\sqrt{3})$ of $R = \mathbb{Z}[\sqrt{-3}]$ lying over $2$. By the same result, the rational prime $2$ is inert in $\widetilde{R} = \mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$, i.e. $\mathfrak{q}=(2)$ is a prime ideal in $\widetilde{R}$. So for the corresponding map of prime spectra $\operatorname{Spec}(\widetilde{R})\rightarrow \operatorname{Spec}(R)$, the preimage of the "singular" point $\mathfrak{p}$ of $\operatorname{Spec}(R)$ consists solely of the "regular" point $\mathfrak{q}$ of $\operatorname{Spec}(\widetilde{R})$. I would therefore say that this situation is more like that of the cuspidal point $(0,0)$ on the singular curve $y^2=x^3$, which also has one point lying over it on the normalization.