I am trying to find some connections between the following two definitions. Let $A$ be a commutative ring with identity and let $\mathfrak p\subset A$ be a prime ideal.
$\mathfrak p$ is a Goldman ideal if there is some $f\in A$ such that $V (\mathfrak p)\cap D(f) =\{\mathfrak p\}$, i.e., the only prime in $D(f)$ containing $\mathfrak p$ is $\mathfrak p$.
Let $\phi: B\to A$ be a finite type ring map. Then $\mathfrak p$ is a isolated prime through $\phi$ if there is some $f\in A$ so that the only prime $\mathfrak q$ in $D(f)\subset \operatorname{Spec}(A)$ with $\phi^{-1}(\mathfrak q) = \phi^{-1}(\mathfrak p)$ is $\mathfrak q = \mathfrak p$.
Given a Goldman ideal $\mathfrak p\subset A$, I am guessing it should be a isolated prime through some finite type ring map $\phi$(Strangely we are constructing a map left to the ring $A$.) The map should satisfy $\phi^{-1}(\mathfrak q) = \phi^{-1}(\mathfrak p)$ iff $\mathfrak q\subset\mathfrak p$. How should we construct the map?
As a reference since the definition for isolated prime is usually different from 2., see equivalent condition (3) here on Stacks Project.