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?
1 Answer
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.
See Korollar 2.12 in
E. Brieskorn: Rationale Singularitäten komplexer Flächen, Invent. Math. 4, 336358 (1968). ZBL0219.14003.

$\begingroup$ 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 $\endgroup$ Nov 17, 2021 at 6:20

$\begingroup$ 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. $\endgroup$ Nov 17, 2021 at 6:53

$\begingroup$ Right sorry, I misunderstood that point. But here's another question: By uniquely characterized by "resolution graph", do we mean the resolution graph with the correct weight of selfintersection number at each vertex? $\endgroup$ Nov 17, 2021 at 23:52

$\begingroup$ Yes, the weighted resolution graph. $\endgroup$ Nov 18, 2021 at 7:26