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Added a reference. To the OP: if you know a better one, please change.
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მამუკა ჯიბლაძე
მამუკა ჯიბლაძე
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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 singularitycyclic quotient singularity?

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?

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?

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Snake Eyes
Snake Eyes
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$2$-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?