为何曲线到射影空间的映射可在正则闭点处唯一延拓？及相关事实疑问

Question 1: Why is the extension of a curve-to-projective-space morphism unique at a regular closed point? #

First, let's anchor this to the theorem you cited (Vakil's 16.5.1): Suppose $C$ is a pure 1-dimensional Noetherian scheme over an affine base $S$, $p \in C$ is a regular closed point, and $Y$ is a projective $S$-scheme. Any morphism $C \setminus {p} \to Y$ extends uniquely to all of $C$.

The uniqueness argument has two complementary angles, as Vakil outlines:

1. Reduced neighborhood trick:
   Since $p$ is regular, its local ring $\mathscr{O}_{C,p}$ is a 1-dimensional regular local ring. A core property of regular local rings is that they're integral domains (hence reduced—no nilpotent elements). This means we can find an open neighborhood of $p$ that's reduced. Now, if two extensions $f, g: C \to Y$ agree on $C \setminus {p}$, their restrictions to this reduced neighborhood must coincide everywhere. Why? For reduced schemes, any two morphisms that agree on a dense open subset are identical—because the set where they differ would require a nilpotent ideal, which can't exist in a reduced scheme. Since $C \setminus {p}$ is dense in this neighborhood, $f$ and $g$ must be the same here, and thus on all of $C$.

2. Separation of projective schemes:
   Projective $S$-schemes are separated (a consequence of properness, which projective implies). For separated schemes, the equalizer of two morphisms $f, g: C \to Y$ is a closed subscheme of $C$. We know $C \setminus {p}$ is contained in this equalizer, so the complement of the equalizer is a closed subset contained in ${p}$. Now, since $p$ is regular, ${p}$ is an effective Cartier divisor (more on this in question 2), which means $C \setminus {p}$ is dense in $C$. A closed subscheme containing a dense open subset must be the entire scheme—so the equalizer is all of $C$, hence $f = g$.

Question 2: Why is $\mathscr{O}_{C,p}$ reduced, or why is $p$ cut out by a non-zero divisor? #

Let's unpack the setup: $p$ is a regular closed point of a pure 1-dimensional Noetherian scheme $C$.

• $\mathscr{O}_{C,p}$ is reduced:
  A regular local ring of dimension 1 is an integral domain. This is a standard commutative algebra result: regular local rings are unique factorization domains (UFDs), and UFDs are integral domains (no zero divisors, hence reduced). Since $\mathscr{O}_{C,p}$ is regular (by definition of $p$ being a regular point) and dimension 1 (since $C$ is pure 1-dimensional), it's an integral domain, so reduced.

• $p$ is cut out by a non-zero divisor:
  For a 1-dimensional regular local ring $\mathscr{O}_{C,p}$, its maximal ideal $\mathfrak{m}p$ is generated by a single element $t$ (this is exactly what "regular" means here: the minimal number of generators of the maximal ideal equals the ring's dimension). Since $\mathscr{O}{C,p}$ is an integral domain, $t \neq 0$, and in an integral domain, every non-zero element is a non-zero divisor. Therefore, the ideal defining the closed point $p$ (locally) is $(t)$, generated by a non-zero divisor—so ${p}$ is an effective Cartier divisor near $p$.

These two facts are critical for the uniqueness argument, as they ensure removing $p$ doesn't erase essential structural information about the scheme, forcing the extension to be one-of-a-kind.

内容的提问来源于stack exchange，提问作者Aaron Johnson