$\operatorname{Spec}(A[\frac{b}{a}]) \longrightarrow \operatorname{Spec}(A)$ is an open immersion when $A$ is a Dedekind domain.

Question

Let $A$ be a Dedekind domain, $a, b \in A$ with $a \neq 0$. I wonder if there is an elegant proof (using properties of Dedekind domain, not Zariski's main theorem of course) to show that $\operatorname{Spec}(A[\frac{b}{a}]) \longrightarrow \operatorname{Spec}(A)$ is an open immersion.

Background:

This is (affine case of) exercise 4.1.14 in Liu Qing's book Algebraic Geometry and Arithmetic curves.

EGA 4(3), Corollary 8.12.10

Soient Y un préschéma intègre et normal, X un préschéma intègre, $f: X \longrightarrow Y$ un morphisme birationnel et localement quasi-fini (Errm , 20). Alors $f$ est un isomorphisme local ; pour que $f$ soit une immersion ouverte, il faut et il suffit que $f$ soit en outre séparé.

provides a general proposition which is itself a simple corollary of Zariski's main theorem.

The purpose of this exercise is (I guess) to examine the dimension one (Dedekind domain) case ''by hands''.