Let $X$ be a scheme over a field $k$. Let $X = X_1 \cup \cdots \cup X_n$ be its decomposition into irreducible components. If a point $x \in X$ lies in more than one component, is it necessarily singular? Why?
Points lying on more than one irreducible component
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algebraic-geometry
commutative-algebra
1 Answers
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Let me assume that $X$ is locally Noetherian; otherwise, I don't know what the word “singular” means. The minimal primes of the local ring $\mathcal O_{X, x}$ correspond to the irreducible components of $X$ passing through $x$. If there is more than one of these then $\mathcal O_{X, x}$ is certainly not a domain, and it is a fact that regular local rings are domains.
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0I second that, really nice answer! – 2012-09-05