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The curve-to-projective extension theorem says:

If $C$ is a pure dimension 1 Noetherian scheme over an affine base $S$, and $p \in C$ is a regular closed point of it. Suppose $Y$ is a projective $S$-scheme. Then any morphism $C \setminus \{p\} \to Y$ extends to all of $C$.

This is 16.5.1 in Vakil's notes. Then he argues that this extension is unique in two ways: one is to replace $C$ by an open neighborhood of $p$ that is reduced, and the other is to note that maps to separated schemes can be extended over an effective Cartier divisor in at most one way.

For the argument to go through we need the fact that $\mathscr{O}_{C,p}$ is reduced, or that $p$ is cut out by a non-zero-divisor. Why is that true?

Elías Guisado Villalgordo
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Aaron Johnson
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Since $p$ is a regular point, $\mathscr{O}_{C, p}$ is a regular local ring, which is reduced. The nonreduced locus of $C$ is the closure of the associated points whose stalks are nonreduced, therefore is closed. So there's an open neighborhood $U$ of $p$ where $C$ is reduced. $Y$ is projective, therefore separated, and a map from a reduced scheme to a separated scheme is determined by its behavior on a dense open set. So the map $U \to Y$ is determined by $U \setminus\{p\} \to Y$.

For the other argument, since $p$ is a regular point, the maximal ideal of $\mathscr{O}_{C,p}$ is principal (a dimension 1 Noetherian local ring is regular if and only if its maximal ideal is principal, this is e.g. theorem 12.5.1 in Vakil), and is generated by a nonzerodivisor since $\mathscr{O}_{C,p}$ is a domain, so $\{p\}$ is an effective Cartier divisor on $C$.

Aaron Johnson
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