# $2$-dimensional complete local normal domain with rational singularity that has exactly one exceptional curve

Let $$(R, \mathfrak m)$$ be a complete local normal domain of dimension $$2$$ with residue field $$R/\mathfrak m$$ algebraically closed and characteristic $$0$$. Assume Spec$$(R)$$ has rational singularity, let $$\pi: X \to \text{Spec}(R)$$ be minimal resolution of singularities with exceptional divisor $$E=\pi^{-1}(\mathfrak m)$$. If $$E$$ has exactly one irreducible component, then must it be true that $$R$$ is a cyclic quotient singularity?

The answer is yes, at least over $$\mathbb{C}$$, since $$2$$-dimensional (cyclic) quotient singularities are taut (starr, in German), namely, they are uniquely characterized, up to biholomorphisms, by their resolution graph.
In other words, every $$2$$-dimensional normal singularity, having the same resolution graph of a (cyclic) quotient singularity, is itself a (cyclic) quotient singularity.
• Sorry, but I am not sure why $R$ even has to be a quotient singularity to begin with? Could you please explain this point? Thanks Nov 17 '21 at 6:20
• I am not assuming a priori that $R$ is a quotient singularity, but only that it is a rational normal singularity. Brieskorn's tautness result applies to this situation and implies that, if the resolution graph of $R$ is the same as the graph of a quotient singularity, then $R$ is actually a quotient singularity. See also the definition at p. 349 of the linked paper. Nov 17 '21 at 6:53